Solve 7x^2-4left(x^2+x+2right)=3x^2+4x-8

$$7 { x }^{ 2 } -4 \left( { x }^{ 2 } +x+2 \right) = 3 { x }^{ 2 } +4x-8$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

Use the distributive property to multiply -4 by x^{2}+x+2.

$$7x^{2}-4x^{2}-4x-8=3x^{2}+4x-8$$

Combine 7x^{2} and -4x^{2} to get 3x^{2}.

$$3x^{2}-4x-8=3x^{2}+4x-8$$

Subtract 3x^{2} from both sides.

$$3x^{2}-4x-8-3x^{2}=4x-8$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-4x-8=4x-8$$

Subtract 4x from both sides.

$$-4x-8-4x=-8$$

Combine -4x and -4x to get -8x.

$$-8x-8=-8$$

Add 8 to both sides.

$$-8x=-8+8$$

Add -8 and 8 to get 0.

$$-8x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -8 is not equal to 0, x must be equal to 0.

$$x=0$$