Solve 2x^2-4+(6*2)=24

$$2 x ^ { 2 } - 4 + ( 6 \times 2 ) = 24$$

Solve for x

$$x=2\sqrt{2}\approx 2.828427125$$

$$x=-2\sqrt{2}\approx -2.828427125$$

Steps by Finding Square Root

Multiply 6 and 2 to get 12.

$$2x^{2}-4+12=24$$

Add -4 and 12 to get 8.

$$2x^{2}+8=24$$

Subtract 8 from both sides.

$$2x^{2}=24-8$$

Subtract 8 from 24 to get 16.

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

Divide both sides by 2.

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

Divide 16 by 2 to get 8.

$$x^{2}=8$$

Take the square root of both sides of the equation.

$$x=2\sqrt{2} x=-2\sqrt{2}$$

Steps Using the Quadratic Formula

Multiply 6 and 2 to get 12.

$$2x^{2}-4+12=24$$

Add -4 and 12 to get 8.

$$2x^{2}+8=24$$

Subtract 24 from both sides.

$$2x^{2}+8-24=0$$

Subtract 24 from 8 to get -16.

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

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

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

Square 0.

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

Multiply -4 times 2.

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

Multiply -8 times -16.

$$x=\frac{0±\sqrt{128}}{2\times 2}$$

Take the square root of 128.

$$x=\frac{0±8\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±8\sqrt{2}}{4}$$

Now solve the equation x=\frac{0±8\sqrt{2}}{4} when ± is plus.

$$x=2\sqrt{2}$$

Now solve the equation x=\frac{0±8\sqrt{2}}{4} when ± is minus.

$$x=-2\sqrt{2}$$

The equation is now solved.

$$x=2\sqrt{2} x=-2\sqrt{2}$$