2x 2 16x 14 0

$2 \exponential{x}{ii} - 16 ten + xiv $

two\left(x-7\correct)\left(ten-1\correct)

2\left(x-7\correct)\left(10-1\right)

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ii\left(ten^{2}-8x+7\right)

Factor out 2.

a+b=-8 ab=ane\times vii=7

Consider ten^{2}-8x+7. Factor the expression by grouping. First, the expression needs to exist rewritten as x^{2}+ax+bx+7. To detect a and b, set upwardly a organization to exist solved.

a=-7 b=-1

Since ab is positive, a and b take the aforementioned sign. Since a+b is negative, a and b are both negative. The just such pair is the system solution.

\left(x^{ii}-7x\correct)+\left(-x+7\right)

Rewrite x^{2}-8x+vii equally \left(x^{two}-7x\correct)+\left(-ten+seven\right).

x\left(x-7\right)-\left(x-7\right)

Factor out 10 in the first and -1 in the 2nd grouping.

\left(x-7\right)\left(10-one\right)

Factor out common term x-7 past using distributive property.

2\left(10-seven\right)\left(x-ane\right)

Rewrite the consummate factored expression.

2x^{2}-16x+xiv=0

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(10-x_{1}\correct)\left(x-x_{2}\correct), where x_{ane} and x_{two} are the solutions of the quadratic equation ax^{two}+bx+c=0.

x=\frac{-\left(-16\right)±\sqrt{\left(-xvi\right)^{ii}-iv\times 2\times xiv}}{ii\times 2}

All equations of the course ax^{ii}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, i when ± is addition and 1 when it is subtraction.

x=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 14}}{2\times 2}

Square -16.

x=\frac{-\left(-16\right)±\sqrt{256-8\times 14}}{2\times ii}

Multiply -4 times ii.

x=\frac{-\left(-16\right)±\sqrt{256-112}}{2\times 2}

Multiply -8 times 14.

10=\frac{-\left(-16\right)±\sqrt{144}}{2\times 2}

Add 256 to -112.

x=\frac{-\left(-16\right)±12}{2\times two}

Take the square root of 144.

x=\frac{16±12}{2\times ii}

The contrary of -16 is xvi.

x=\frac{xvi±12}{4}

Multiply 2 times two.

x=\frac{28}{iv}

At present solve the equation ten=\frac{sixteen±12}{iv} when ± is plus. Add together sixteen to 12.

x=\frac{iv}{four}

Now solve the equation x=\frac{xvi±12}{four} when ± is minus. Subtract 12 from xvi.

2x^{2}-16x+14=ii\left(ten-7\right)\left(x-1\right)

Factor the original expression using ax^{ii}+bx+c=a\left(10-x_{1}\correct)\left(x-x_{ii}\right). Substitute 7 for x_{1} and 1 for x_{2}.

x ^ ii -8x +7 = 0

Quadratic equations such every bit this ane can be solved by a new direct factoring method that does not require guess work. To utilise the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

r + s = eight rs = 7

Allow r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−south) where sum of factors (r+s)=−B and the product of factors rs = C

r = 4 - u s = 4 + u

Two numbers r and s sum up to viii exactly when the average of the two numbers is \frac{1}{two}*eight = four. You lot can also see that the midpoint of r and south corresponds to the axis of symmetry of the parabola represented past the quadratic equation y=ten^2+Bx+C. The values of r and s are equidistant from the heart by an unknown quantity u. Express r and s with respect to variable u. <div fashion='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' mode='width: 100%;max-width: 700px' /></div>

(4 - u) (iv + u) = 7

To solve for unknown quantity u, substitute these in the product equation rs = vii

xvi - u^2 = 7

Simplify by expanding (a -b) (a + b) = a^2 – b^two

-u^2 = 7-16 = -9

Simplify the expression past subtracting xvi on both sides

u^2 = 9 u = \pm\sqrt{nine} = \pm 3

Simplify the expression by multiplying -1 on both sides and take the foursquare root to obtain the value of unknown variable u

r =iv - three = 1 s = 4 + iii = 7

The factors r and southward are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

2x 2 16x 14 0,

Source: https://mathsolver.microsoft.com/en/solve-problem/2%20x%20%5E%20%7B%202%20%7D%20-%2016%20x%20%2B%2014

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