Solve 2x^2-7x-2=4x+5

$$2 x ^ { 2 } - 7 x - 2 = 4 x + 5$$

Solve for x

$$x = \frac{\sqrt{177} + 11}{4} \approx 6.076033674$$

$$x=\frac{11-\sqrt{177}}{4}\approx -0.576033674$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$2x^{2}-7x-2-4x=5$$

Combine -7x and -4x to get -11x.

$$2x^{2}-11x-2=5$$

Subtract 5 from both sides.

$$2x^{2}-11x-2-5=0$$

Subtract 5 from -2 to get -7.

$$2x^{2}-11x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-7\right)}}{2\times 2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-7\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\left(-7\right)}}{2\times 2}$$

Multiply -8 times -7.

$$x=\frac{-\left(-11\right)±\sqrt{121+56}}{2\times 2}$$

Add 121 to 56.

$$x=\frac{-\left(-11\right)±\sqrt{177}}{2\times 2}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{177}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±\sqrt{177}}{4}$$

Now solve the equation x=\frac{11±\sqrt{177}}{4} when ± is plus. Add 11 to \sqrt{177}.

$$x=\frac{\sqrt{177}+11}{4}$$

Now solve the equation x=\frac{11±\sqrt{177}}{4} when ± is minus. Subtract \sqrt{177} from 11.

$$x=\frac{11-\sqrt{177}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{177}+11}{4} x=\frac{11-\sqrt{177}}{4}$$

Steps for Completing the Square

Subtract 4x from both sides.

$$2x^{2}-7x-2-4x=5$$

Combine -7x and -4x to get -11x.

$$2x^{2}-11x-2=5$$

Add 2 to both sides.

$$2x^{2}-11x=5+2$$

Add 5 and 2 to get 7.

$$2x^{2}-11x=7$$

Divide both sides by 2.

$$\frac{2x^{2}-11x}{2}=\frac{7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{11}{2}x=\frac{7}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=\frac{7}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{7}{2}+\frac{121}{16}$$

Add \frac{7}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{177}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{177}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{177}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{\sqrt{177}}{4} x-\frac{11}{4}=-\frac{\sqrt{177}}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=\frac{\sqrt{177}+11}{4} x=\frac{11-\sqrt{177}}{4}$$