Solve 6x^2-5x=35-14x

$$6 x ^ { 2 } - 5 x = 35 - 14 x$$

Solve for x

$$x=\frac{\sqrt{921}}{12}-\frac{3}{4}\approx 1.778998484$$

$$x=-\frac{\sqrt{921}}{12}-\frac{3}{4}\approx -3.278998484$$

Steps Using the Quadratic Formula

Subtract 35 from both sides.

$$6x^{2}-5x-35=-14x$$

Add 14x to both sides.

$$6x^{2}-5x-35+14x=0$$

Combine -5x and 14x to get 9x.

$$6x^{2}+9x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 9 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 6\left(-35\right)}}{2\times 6}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 6\left(-35\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-9±\sqrt{81-24\left(-35\right)}}{2\times 6}$$

Multiply -24 times -35.

$$x=\frac{-9±\sqrt{81+840}}{2\times 6}$$

Add 81 to 840.

$$x=\frac{-9±\sqrt{921}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-9±\sqrt{921}}{12}$$

Now solve the equation x=\frac{-9±\sqrt{921}}{12} when ± is plus. Add -9 to \sqrt{921}.

$$x=\frac{\sqrt{921}-9}{12}$$

Divide -9+\sqrt{921} by 12.

$$x=\frac{\sqrt{921}}{12}-\frac{3}{4}$$

Now solve the equation x=\frac{-9±\sqrt{921}}{12} when ± is minus. Subtract \sqrt{921} from -9.

$$x=\frac{-\sqrt{921}-9}{12}$$

Divide -9-\sqrt{921} by 12.

$$x=-\frac{\sqrt{921}}{12}-\frac{3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{921}}{12}-\frac{3}{4} x=-\frac{\sqrt{921}}{12}-\frac{3}{4}$$

Steps for Completing the Square

Add 14x to both sides.

$$6x^{2}-5x+14x=35$$

Combine -5x and 14x to get 9x.

$$6x^{2}+9x=35$$

Divide both sides by 6.

$$\frac{6x^{2}+9x}{6}=\frac{35}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{9}{6}x=\frac{35}{6}$$

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{3}{2}x=\frac{35}{6}$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{35}{6}+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{35}{6}+\frac{9}{16}$$

Add \frac{35}{6} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{307}{48}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{307}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{307}{48}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{921}}{12} x+\frac{3}{4}=-\frac{\sqrt{921}}{12}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{921}}{12}-\frac{3}{4} x=-\frac{\sqrt{921}}{12}-\frac{3}{4}$$