Solve left(5-xright)^2+7=4x^2

$${ \left(5-x \right) }^{ 2 } +7 = 4 { x }^{ 2 }$$

Solve for x

$$x = -\frac{16}{3} = -5\frac{1}{3} \approx -5.333333333$$

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$25-10x+x^{2}+7=4x^{2}$$

Add 25 and 7 to get 32.

$$32-10x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$32-10x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$32-10x-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-10x+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+32. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-3\times 32=-96$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -10.

$$a=6 b=-16$$

Rewrite -3x^{2}-10x+32 as \left(-3x^{2}+6x\right)+\left(-16x+32\right).

$$\left(-3x^{2}+6x\right)+\left(-16x+32\right)$$

Factor out 3x in the first and 16 in the second group.

$$3x\left(-x+2\right)+16\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(3x+16\right)$$

To find equation solutions, solve -x+2=0 and 3x+16=0.

$$x=2 x=-\frac{16}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$25-10x+x^{2}+7=4x^{2}$$

Add 25 and 7 to get 32.

$$32-10x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$32-10x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$32-10x-3x^{2}=0$$

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

$$-3x^{2}-10x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -10 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)\times 32}}{2\left(-3\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)\times 32}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+12\times 32}}{2\left(-3\right)}$$

Multiply 12 times 32.

$$x=\frac{-\left(-10\right)±\sqrt{100+384}}{2\left(-3\right)}$$

Add 100 to 384.

$$x=\frac{-\left(-10\right)±\sqrt{484}}{2\left(-3\right)}$$

Take the square root of 484.

$$x=\frac{-\left(-10\right)±22}{2\left(-3\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±22}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{10±22}{-6}$$

Now solve the equation x=\frac{10±22}{-6} when ± is plus. Add 10 to 22.

$$x=\frac{32}{-6}$$

Reduce the fraction \frac{32}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{16}{3}$$

Now solve the equation x=\frac{10±22}{-6} when ± is minus. Subtract 22 from 10.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

$$x=-\frac{16}{3} x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$25-10x+x^{2}+7=4x^{2}$$

Add 25 and 7 to get 32.

$$32-10x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$32-10x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$32-10x-3x^{2}=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$-10x-3x^{2}=-32$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3x^{2}-10x=-32$$

Divide both sides by -3.

$$\frac{-3x^{2}-10x}{-3}=-\frac{32}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{10}{-3}\right)x=-\frac{32}{-3}$$

Divide -10 by -3.

$$x^{2}+\frac{10}{3}x=-\frac{32}{-3}$$

Divide -32 by -3.

$$x^{2}+\frac{10}{3}x=\frac{32}{3}$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{32}{3}+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{32}{3}+\frac{25}{9}$$

Add \frac{32}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{121}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{121}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{11}{3} x+\frac{5}{3}=-\frac{11}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=2 x=-\frac{16}{3}$$