Solve left(3xright)^2+6^2=left(5xright)^2

$${ \left(3x \right) }^{ 2 } + { 6 }^{ 2 } = { \left(5x \right) }^{ 2 }$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps by Finding Square Root

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}+6^{2}=\left(5x\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}+6^{2}=\left(5x\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$9x^{2}+36=\left(5x\right)^{2}$$

Expand \left(5x\right)^{2}.

$$9x^{2}+36=5^{2}x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$9x^{2}+36=25x^{2}$$

Subtract 25x^{2} from both sides.

$$9x^{2}+36-25x^{2}=0$$

Combine 9x^{2} and -25x^{2} to get -16x^{2}.

$$-16x^{2}+36=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$-16x^{2}=-36$$

Divide both sides by -16.

$$x^{2}=\frac{-36}{-16}$$

Reduce the fraction \frac{-36}{-16} to lowest terms by extracting and canceling out -4.

$$x^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$x=\frac{3}{2} x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}+6^{2}=\left(5x\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}+6^{2}=\left(5x\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$9x^{2}+36=\left(5x\right)^{2}$$

Expand \left(5x\right)^{2}.

$$9x^{2}+36=5^{2}x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$9x^{2}+36=25x^{2}$$

Subtract 25x^{2} from both sides.

$$9x^{2}+36-25x^{2}=0$$

Combine 9x^{2} and -25x^{2} to get -16x^{2}.

$$-16x^{2}+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 0 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-16\right)\times 36}}{2\left(-16\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-16\right)\times 36}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{0±\sqrt{64\times 36}}{2\left(-16\right)}$$

Multiply 64 times 36.

$$x=\frac{0±\sqrt{2304}}{2\left(-16\right)}$$

Take the square root of 2304.

$$x=\frac{0±48}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{0±48}{-32}$$

Now solve the equation x=\frac{0±48}{-32} when ± is plus. Reduce the fraction \frac{48}{-32} to lowest terms by extracting and canceling out 16.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{0±48}{-32} when ± is minus. Reduce the fraction \frac{-48}{-32} to lowest terms by extracting and canceling out 16.

$$x=\frac{3}{2}$$

The equation is now solved.

$$x=-\frac{3}{2} x=\frac{3}{2}$$