Solve -4(x)=2x^2+7x-4

$$-4(x)=2 { x }^{ 2 } +7x-4$$

Solve for x

$$x=\frac{3\sqrt{17}-11}{4}\approx 0.342329219$$

$$x=\frac{-3\sqrt{17}-11}{4}\approx -5.842329219$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Subtract 7x from both sides.

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

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

$$-11x-2x^{2}=-4$$

Add 4 to both sides.

$$-11x-2x^{2}+4=0$$

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.

$$-2x^{2}-11x+4=0$$

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

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

Square -11.

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

Multiply -4 times -2.

$$x=\frac{-\left(-11\right)±\sqrt{121+8\times 4}}{2\left(-2\right)}$$

Multiply 8 times 4.

$$x=\frac{-\left(-11\right)±\sqrt{121+32}}{2\left(-2\right)}$$

Add 121 to 32.

$$x=\frac{-\left(-11\right)±\sqrt{153}}{2\left(-2\right)}$$

Take the square root of 153.

$$x=\frac{-\left(-11\right)±3\sqrt{17}}{2\left(-2\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±3\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{11±3\sqrt{17}}{-4}$$

Now solve the equation x=\frac{11±3\sqrt{17}}{-4} when ± is plus. Add 11 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}+11}{-4}$$

Divide 11+3\sqrt{17} by -4.

$$x=\frac{-3\sqrt{17}-11}{4}$$

Now solve the equation x=\frac{11±3\sqrt{17}}{-4} when ± is minus. Subtract 3\sqrt{17} from 11.

$$x=\frac{11-3\sqrt{17}}{-4}$$

Divide 11-3\sqrt{17} by -4.

$$x=\frac{3\sqrt{17}-11}{4}$$

The equation is now solved.

$$x=\frac{-3\sqrt{17}-11}{4} x=\frac{3\sqrt{17}-11}{4}$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Subtract 7x from both sides.

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

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

$$-11x-2x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}-11x=-4$$

Divide both sides by -2.

$$\frac{-2x^{2}-11x}{-2}=-\frac{4}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -11 by -2.

$$x^{2}+\frac{11}{2}x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x^{2}+\frac{11}{2}x=2$$

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.

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

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

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

Add 2 to \frac{121}{16}.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{153}{16}$$

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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{153}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{153}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{3\sqrt{17}}{4} x+\frac{11}{4}=-\frac{3\sqrt{17}}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

$$x=\frac{3\sqrt{17}-11}{4} x=\frac{-3\sqrt{17}-11}{4}$$