Solve 16x^2+40x+25=4x^2+40x+100

$$16 { x }^{ 2 } +40x+25=4 { x }^{ 2 } +40x+100$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Difference of Squares

Subtract 4x^{2} from both sides.

$$16x^{2}+40x+25-4x^{2}=40x+100$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}+40x+25=40x+100$$

Subtract 40x from both sides.

$$12x^{2}+40x+25-40x=100$$

Combine 40x and -40x to get 0.

$$12x^{2}+25=100$$

Subtract 100 from both sides.

$$12x^{2}+25-100=0$$

Subtract 100 from 25 to get -75.

$$12x^{2}-75=0$$

Divide both sides by 3.

$$4x^{2}-25=0$$

Consider 4x^{2}-25. Rewrite 4x^{2}-25 as \left(2x\right)^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2x-5\right)\left(2x+5\right)=0$$

To find equation solutions, solve 2x-5=0 and 2x+5=0.

$$x=\frac{5}{2} x=-\frac{5}{2}$$

Steps by Finding Square Root

Subtract 4x^{2} from both sides.

$$16x^{2}+40x+25-4x^{2}=40x+100$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}+40x+25=40x+100$$

Subtract 40x from both sides.

$$12x^{2}+40x+25-40x=100$$

Combine 40x and -40x to get 0.

$$12x^{2}+25=100$$

Subtract 25 from both sides.

$$12x^{2}=100-25$$

Subtract 25 from 100 to get 75.

$$12x^{2}=75$$

Divide both sides by 12.

$$x^{2}=\frac{75}{12}$$

Reduce the fraction \frac{75}{12} to lowest terms by extracting and canceling out 3.

$$x^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{5}{2} x=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$16x^{2}+40x+25-4x^{2}=40x+100$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}+40x+25=40x+100$$

Subtract 40x from both sides.

$$12x^{2}+40x+25-40x=100$$

Combine 40x and -40x to get 0.

$$12x^{2}+25=100$$

Subtract 100 from both sides.

$$12x^{2}+25-100=0$$

Subtract 100 from 25 to get -75.

$$12x^{2}-75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 0 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 12\left(-75\right)}}{2\times 12}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 12\left(-75\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{0±\sqrt{-48\left(-75\right)}}{2\times 12}$$

Multiply -48 times -75.

$$x=\frac{0±\sqrt{3600}}{2\times 12}$$

Take the square root of 3600.

$$x=\frac{0±60}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{0±60}{24}$$

Now solve the equation x=\frac{0±60}{24} when ± is plus. Reduce the fraction \frac{60}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{0±60}{24} when ± is minus. Reduce the fraction \frac{-60}{24} to lowest terms by extracting and canceling out 12.

$$x=-\frac{5}{2}$$

The equation is now solved.

$$x=\frac{5}{2} x=-\frac{5}{2}$$