Solve 68x^2=120-33sqrt{15}

$$68 { x }^{ 2 } =120-33 \sqrt{ 15 }$$

Solve for x (complex solution)

$$x=-\frac{i\sqrt{561\sqrt{15}-2040}}{34}\approx -0-0.338865981i$$

$$x=\frac{i\sqrt{561\sqrt{15}-2040}}{34}\approx 0.338865981i$$

Steps by Finding Square Root

Dividing by 68 undoes the multiplication by 68.

$$x^{2}=\frac{120-33\sqrt{15}}{68}$$

Divide 120-33\sqrt{15} by 68.

$$x^{2}=-\frac{33\sqrt{15}}{68}+\frac{30}{17}$$

Take the square root of both sides of the equation.

$$x=\frac{i\sqrt{561\sqrt{15}-2040}}{34} x=-\frac{i\sqrt{561\sqrt{15}-2040}}{34}$$

Steps Using the Quadratic Formula

Subtract 120 from both sides.

$$68x^{2}-120=-33\sqrt{15}$$

Add 33\sqrt{15} to both sides.

$$68x^{2}-120+33\sqrt{15}=0$$

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.

$$68x^{2}+33\sqrt{15}-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 68 for a, 0 for b, and -120+33\sqrt{15} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 68\left(33\sqrt{15}-120\right)}}{2\times 68}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 68\left(33\sqrt{15}-120\right)}}{2\times 68}$$

Multiply -4 times 68.

$$x=\frac{0±\sqrt{-272\left(33\sqrt{15}-120\right)}}{2\times 68}$$

Multiply -272 times -120+33\sqrt{15}.

$$x=\frac{0±\sqrt{32640-8976\sqrt{15}}}{2\times 68}$$

Take the square root of 32640-8976\sqrt{15}.

$$x=\frac{0±4i\sqrt{561\sqrt{15}-2040}}{2\times 68}$$

Multiply 2 times 68.

$$x=\frac{0±4i\sqrt{561\sqrt{15}-2040}}{136}$$

Now solve the equation x=\frac{0±4i\sqrt{561\sqrt{15}-2040}}{136} when ± is plus.

$$x=\frac{i\sqrt{561\sqrt{15}-2040}}{34}$$

Now solve the equation x=\frac{0±4i\sqrt{561\sqrt{15}-2040}}{136} when ± is minus.

$$x=-\frac{i\sqrt{561\sqrt{15}-2040}}{34}$$

The equation is now solved.

$$x=\frac{i\sqrt{561\sqrt{15}-2040}}{34} x=-\frac{i\sqrt{561\sqrt{15}-2040}}{34}$$