Solve 36x^2+84x=-49

$$36 x ^ { 2 } + 84 x = - 49$$

Solve for x

$$x = -\frac{7}{6} = -1\frac{1}{6} \approx -1.166666667$$

Steps Using the Quadratic Formula

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.

$$36x^{2}+84x=-49$$

Add 49 to both sides of the equation.

$$36x^{2}+84x-\left(-49\right)=-49-\left(-49\right)$$

Subtracting -49 from itself leaves 0.

$$36x^{2}+84x-\left(-49\right)=0$$

Subtract -49 from 0.

$$36x^{2}+84x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 84 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-84±\sqrt{84^{2}-4\times 36\times 49}}{2\times 36}$$

Square 84.

$$x=\frac{-84±\sqrt{7056-4\times 36\times 49}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-84±\sqrt{7056-144\times 49}}{2\times 36}$$

Multiply -144 times 49.

$$x=\frac{-84±\sqrt{7056-7056}}{2\times 36}$$

Add 7056 to -7056.

$$x=\frac{-84±\sqrt{0}}{2\times 36}$$

Take the square root of 0.

$$x=-\frac{84}{2\times 36}$$

Multiply 2 times 36.

$$x=-\frac{84}{72}$$

Reduce the fraction \frac{-84}{72} to lowest terms by extracting and canceling out 12.

$$x=-\frac{7}{6}$$

Steps for Completing the Square

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.

$$36x^{2}+84x=-49$$

Divide both sides by 36.

$$\frac{36x^{2}+84x}{36}=-\frac{49}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\frac{84}{36}x=-\frac{49}{36}$$

Reduce the fraction \frac{84}{36} to lowest terms by extracting and canceling out 12.

$$x^{2}+\frac{7}{3}x=-\frac{49}{36}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-\frac{49}{36}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{-49+49}{36}$$

Add -\frac{49}{36} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=0$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+\frac{7}{6}=0 x+\frac{7}{6}=0$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=-\frac{7}{6} x=-\frac{7}{6}$$

The equation is now solved. Solutions are the same.

$$x=-\frac{7}{6}$$