Solve 2x^2-5=-4x

$$2 x ^ { 2 } - 5 = - 4 x$$

Solve for x

$$x=\frac{\sqrt{14}}{2}-1\approx 0.870828693$$

$$x=-\frac{\sqrt{14}}{2}-1\approx -2.870828693$$

Steps Using the Quadratic Formula

Add 4x to both sides.

$$2x^{2}-5+4x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}+4x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-4±\sqrt{16+40}}{2\times 2}$$

Add 16 to 40.

$$x=\frac{-4±\sqrt{56}}{2\times 2}$$

Take the square root of 56.

$$x=\frac{-4±2\sqrt{14}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±2\sqrt{14}}{4}$$

Now solve the equation x=\frac{-4±2\sqrt{14}}{4} when ± is plus. Add -4 to 2\sqrt{14}.

$$x=\frac{2\sqrt{14}-4}{4}$$

Divide -4+2\sqrt{14} by 4.

$$x=\frac{\sqrt{14}}{2}-1$$

Now solve the equation x=\frac{-4±2\sqrt{14}}{4} when ± is minus. Subtract 2\sqrt{14} from -4.

$$x=\frac{-2\sqrt{14}-4}{4}$$

Divide -4-2\sqrt{14} by 4.

$$x=-\frac{\sqrt{14}}{2}-1$$

The equation is now solved.

$$x=\frac{\sqrt{14}}{2}-1 x=-\frac{\sqrt{14}}{2}-1$$

Steps for Completing the Square

Add 4x to both sides.

$$2x^{2}-5+4x=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$2x^{2}+4x=5$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=\frac{5}{2}$$

Divide 4 by 2.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{5}{2}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{5}{2}+1$$

Add \frac{5}{2} to 1.

$$x^{2}+2x+1=\frac{7}{2}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{7}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{7}{2}}$$

Simplify.

$$x+1=\frac{\sqrt{14}}{2} x+1=-\frac{\sqrt{14}}{2}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{14}}{2}-1 x=-\frac{\sqrt{14}}{2}-1$$