Solve (2x-3)+(x+5)^2+[(2x+3)-(x-5)]^2=10x^2+92

$$( 2 x - 3 ) + ( x + 5 ) ^ { 2 } + [ ( 2 x + 3 ) - ( x - 5 ) ] ^ { 2 } = 10 x ^ { 2 } + 92$$

Solve for x

$$x = \frac{\sqrt{37} + 7}{4} \approx 3.270690633$$

$$x=\frac{7-\sqrt{37}}{4}\approx 0.229309367$$

Steps Using the Quadratic Formula

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

$$2x-3+x^{2}+10x+25+\left(2x+3-\left(x-5\right)\right)^{2}=10x^{2}+92$$

Combine 2x and 10x to get 12x.

$$12x-3+x^{2}+25+\left(2x+3-\left(x-5\right)\right)^{2}=10x^{2}+92$$

Add -3 and 25 to get 22.

$$12x+22+x^{2}+\left(2x+3-\left(x-5\right)\right)^{2}=10x^{2}+92$$

To find the opposite of x-5, find the opposite of each term.

$$12x+22+x^{2}+\left(2x+3-x+5\right)^{2}=10x^{2}+92$$

Combine 2x and -x to get x.

$$12x+22+x^{2}+\left(x+3+5\right)^{2}=10x^{2}+92$$

Add 3 and 5 to get 8.

$$12x+22+x^{2}+\left(x+8\right)^{2}=10x^{2}+92$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+8\right)^{2}.

$$12x+22+x^{2}+x^{2}+16x+64=10x^{2}+92$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$12x+22+2x^{2}+16x+64=10x^{2}+92$$

Combine 12x and 16x to get 28x.

$$28x+22+2x^{2}+64=10x^{2}+92$$

Add 22 and 64 to get 86.

$$28x+86+2x^{2}=10x^{2}+92$$

Subtract 10x^{2} from both sides.

$$28x+86+2x^{2}-10x^{2}=92$$

Combine 2x^{2} and -10x^{2} to get -8x^{2}.

$$28x+86-8x^{2}=92$$

Subtract 92 from both sides.

$$28x+86-8x^{2}-92=0$$

Subtract 92 from 86 to get -6.

$$28x-6-8x^{2}=0$$

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.

$$-8x^{2}+28x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 28 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\left(-8\right)\left(-6\right)}}{2\left(-8\right)}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\left(-8\right)\left(-6\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-28±\sqrt{784+32\left(-6\right)}}{2\left(-8\right)}$$

Multiply 32 times -6.

$$x=\frac{-28±\sqrt{784-192}}{2\left(-8\right)}$$

Add 784 to -192.

$$x=\frac{-28±\sqrt{592}}{2\left(-8\right)}$$

Take the square root of 592.

$$x=\frac{-28±4\sqrt{37}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-28±4\sqrt{37}}{-16}$$

Now solve the equation x=\frac{-28±4\sqrt{37}}{-16} when ± is plus. Add -28 to 4\sqrt{37}.

$$x=\frac{4\sqrt{37}-28}{-16}$$

Divide -28+4\sqrt{37} by -16.

$$x=\frac{7-\sqrt{37}}{4}$$

Now solve the equation x=\frac{-28±4\sqrt{37}}{-16} when ± is minus. Subtract 4\sqrt{37} from -28.

$$x=\frac{-4\sqrt{37}-28}{-16}$$

Divide -28-4\sqrt{37} by -16.

$$x=\frac{\sqrt{37}+7}{4}$$

The equation is now solved.

$$x=\frac{7-\sqrt{37}}{4} x=\frac{\sqrt{37}+7}{4}$$

Steps for Completing the Square

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

$$2x-3+x^{2}+10x+25+\left(2x+3-\left(x-5\right)\right)^{2}=10x^{2}+92$$

Combine 2x and 10x to get 12x.

$$12x-3+x^{2}+25+\left(2x+3-\left(x-5\right)\right)^{2}=10x^{2}+92$$

Add -3 and 25 to get 22.

$$12x+22+x^{2}+\left(2x+3-\left(x-5\right)\right)^{2}=10x^{2}+92$$

To find the opposite of x-5, find the opposite of each term.

$$12x+22+x^{2}+\left(2x+3-x+5\right)^{2}=10x^{2}+92$$

Combine 2x and -x to get x.

$$12x+22+x^{2}+\left(x+3+5\right)^{2}=10x^{2}+92$$

Add 3 and 5 to get 8.

$$12x+22+x^{2}+\left(x+8\right)^{2}=10x^{2}+92$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+8\right)^{2}.

$$12x+22+x^{2}+x^{2}+16x+64=10x^{2}+92$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$12x+22+2x^{2}+16x+64=10x^{2}+92$$

Combine 12x and 16x to get 28x.

$$28x+22+2x^{2}+64=10x^{2}+92$$

Add 22 and 64 to get 86.

$$28x+86+2x^{2}=10x^{2}+92$$

Subtract 10x^{2} from both sides.

$$28x+86+2x^{2}-10x^{2}=92$$

Combine 2x^{2} and -10x^{2} to get -8x^{2}.

$$28x+86-8x^{2}=92$$

Subtract 86 from both sides.

$$28x-8x^{2}=92-86$$

Subtract 86 from 92 to get 6.

$$28x-8x^{2}=6$$

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.

$$-8x^{2}+28x=6$$

Divide both sides by -8.

$$\frac{-8x^{2}+28x}{-8}=\frac{6}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{28}{-8}x=\frac{6}{-8}$$

Reduce the fraction \frac{28}{-8} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{7}{2}x=\frac{6}{-8}$$

Reduce the fraction \frac{6}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{2}x=-\frac{3}{4}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{3}{4}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{3}{4}+\frac{49}{16}$$

Add -\frac{3}{4} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{37}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{37}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{37}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{37}}{4} x-\frac{7}{4}=-\frac{\sqrt{37}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{37}+7}{4} x=\frac{7-\sqrt{37}}{4}$$