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

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

Solve for x

$$x=\frac{\sqrt{177}-7}{8}\approx 0.788016837$$

$$x=\frac{-\sqrt{177}-7}{8}\approx -2.538016837$$

Steps Using the Quadratic Formula

Subtract 2 from -2 to get -4.

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

Subtract 4 from both sides.

$$-4+4x^{2}+7x-4=0$$

Subtract 4 from -4 to get -8.

$$-8+4x^{2}+7x=0$$

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.

$$4x^{2}+7x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 7 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 4\left(-8\right)}}{2\times 4}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 4\left(-8\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-7±\sqrt{49-16\left(-8\right)}}{2\times 4}$$

Multiply -16 times -8.

$$x=\frac{-7±\sqrt{49+128}}{2\times 4}$$

Add 49 to 128.

$$x=\frac{-7±\sqrt{177}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-7±\sqrt{177}}{8}$$

Now solve the equation x=\frac{-7±\sqrt{177}}{8} when ± is plus. Add -7 to \sqrt{177}.

$$x=\frac{\sqrt{177}-7}{8}$$

Now solve the equation x=\frac{-7±\sqrt{177}}{8} when ± is minus. Subtract \sqrt{177} from -7.

$$x=\frac{-\sqrt{177}-7}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{177}-7}{8} x=\frac{-\sqrt{177}-7}{8}$$

Steps for Completing the Square

Subtract 2 from -2 to get -4.

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

Add 4 to both sides.

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

Add 4 and 4 to get 8.

$$4x^{2}+7x=8$$

Divide both sides by 4.

$$\frac{4x^{2}+7x}{4}=\frac{8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{7}{4}x=\frac{8}{4}$$

Divide 8 by 4.

$$x^{2}+\frac{7}{4}x=2$$

Divide \frac{7}{4}, the coefficient of the x term, by 2 to get \frac{7}{8}. Then add the square of \frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{4}x+\left(\frac{7}{8}\right)^{2}=2+\left(\frac{7}{8}\right)^{2}$$

Square \frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=2+\frac{49}{64}$$

Add 2 to \frac{49}{64}.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{177}{64}$$

Factor x^{2}+\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{8}\right)^{2}=\frac{177}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{8}\right)^{2}}=\sqrt{\frac{177}{64}}$$

Simplify.

$$x+\frac{7}{8}=\frac{\sqrt{177}}{8} x+\frac{7}{8}=-\frac{\sqrt{177}}{8}$$

Subtract \frac{7}{8} from both sides of the equation.

$$x=\frac{\sqrt{177}-7}{8} x=\frac{-\sqrt{177}-7}{8}$$