Solve (7-2x)*(7+2x)=(2x+3)^2-12x

$$( 7 - 2 x ) \cdot ( 7 + 2 x ) = ( 2 x + 3 ) ^ { 2 } - 12 x$$

Solve for x

$$x=\sqrt{5}\approx 2.236067977$$

$$x=-\sqrt{5}\approx -2.236067977$$

Steps by Finding Square Root

Consider \left(7-2x\right)\left(7+2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 7.

$$49-\left(2x\right)^{2}=\left(2x+3\right)^{2}-12x$$

Expand \left(2x\right)^{2}.

$$49-2^{2}x^{2}=\left(2x+3\right)^{2}-12x$$

Calculate 2 to the power of 2 and get 4.

$$49-4x^{2}=\left(2x+3\right)^{2}-12x$$

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

$$49-4x^{2}=4x^{2}+12x+9-12x$$

Combine 12x and -12x to get 0.

$$49-4x^{2}=4x^{2}+9$$

Subtract 4x^{2} from both sides.

$$49-4x^{2}-4x^{2}=9$$

Combine -4x^{2} and -4x^{2} to get -8x^{2}.

$$49-8x^{2}=9$$

Subtract 49 from both sides.

$$-8x^{2}=9-49$$

Subtract 49 from 9 to get -40.

$$-8x^{2}=-40$$

Divide both sides by -8.

$$x^{2}=\frac{-40}{-8}$$

Divide -40 by -8 to get 5.

$$x^{2}=5$$

Take the square root of both sides of the equation.

$$x=\sqrt{5} x=-\sqrt{5}$$

Steps Using the Quadratic Formula

Consider \left(7-2x\right)\left(7+2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 7.

$$49-\left(2x\right)^{2}=\left(2x+3\right)^{2}-12x$$

Expand \left(2x\right)^{2}.

$$49-2^{2}x^{2}=\left(2x+3\right)^{2}-12x$$

Calculate 2 to the power of 2 and get 4.

$$49-4x^{2}=\left(2x+3\right)^{2}-12x$$

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

$$49-4x^{2}=4x^{2}+12x+9-12x$$

Combine 12x and -12x to get 0.

$$49-4x^{2}=4x^{2}+9$$

Subtract 4x^{2} from both sides.

$$49-4x^{2}-4x^{2}=9$$

Combine -4x^{2} and -4x^{2} to get -8x^{2}.

$$49-8x^{2}=9$$

Subtract 9 from both sides.

$$49-8x^{2}-9=0$$

Subtract 9 from 49 to get 40.

$$40-8x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-8x^{2}+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 0 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-8\right)\times 40}}{2\left(-8\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-8\right)\times 40}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{0±\sqrt{32\times 40}}{2\left(-8\right)}$$

Multiply 32 times 40.

$$x=\frac{0±\sqrt{1280}}{2\left(-8\right)}$$

Take the square root of 1280.

$$x=\frac{0±16\sqrt{5}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{0±16\sqrt{5}}{-16}$$

Now solve the equation x=\frac{0±16\sqrt{5}}{-16} when ± is plus.

$$x=-\sqrt{5}$$

Now solve the equation x=\frac{0±16\sqrt{5}}{-16} when ± is minus.

$$x=\sqrt{5}$$

The equation is now solved.

$$x=-\sqrt{5} x=\sqrt{5}$$