Solve (2x-1)^2-(3x+4)^2=-5x(x+8)

$$( 2 x - 1 ) ^ { 2 } - ( 3 x + 4 ) ^ { 2 } = - 5 x ( x + 8 )$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1-\left(3x+4\right)^{2}=-5x\left(x+8\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$4x^{2}-4x+1-\left(9x^{2}+24x+16\right)=-5x\left(x+8\right)$$

To find the opposite of 9x^{2}+24x+16, find the opposite of each term.

$$4x^{2}-4x+1-9x^{2}-24x-16=-5x\left(x+8\right)$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+1-24x-16=-5x\left(x+8\right)$$

Combine -4x and -24x to get -28x.

$$-5x^{2}-28x+1-16=-5x\left(x+8\right)$$

Subtract 16 from 1 to get -15.

$$-5x^{2}-28x-15=-5x\left(x+8\right)$$

Use the distributive property to multiply -5x by x+8.

$$-5x^{2}-28x-15=-5x^{2}-40x$$

Add 5x^{2} to both sides.

$$-5x^{2}-28x-15+5x^{2}=-40x$$

Combine -5x^{2} and 5x^{2} to get 0.

$$-28x-15=-40x$$

Add 40x to both sides.

$$-28x-15+40x=0$$

Combine -28x and 40x to get 12x.

$$12x-15=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$12x=15$$

Divide both sides by 12.

$$x=\frac{15}{12}$$

Reduce the fraction \frac{15}{12} to lowest terms by extracting and canceling out 3.

$$x=\frac{5}{4}$$