Solve 6-6x^2-3x=

$$6 - 6 x ^ { 2 } - 3 x =$$

Factor

$$-6\left(x-\frac{-\sqrt{17}-1}{4}\right)\left(x-\frac{\sqrt{17}-1}{4}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-6x^{2}-3x+6=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24\times 6}}{2\left(-6\right)}$$

Multiply 24 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9+144}}{2\left(-6\right)}$$

Add 9 to 144.

$$x=\frac{-\left(-3\right)±\sqrt{153}}{2\left(-6\right)}$$

Take the square root of 153.

$$x=\frac{-\left(-3\right)±3\sqrt{17}}{2\left(-6\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{17}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{3±3\sqrt{17}}{-12}$$

Now solve the equation x=\frac{3±3\sqrt{17}}{-12} when ± is plus. Add 3 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}+3}{-12}$$

Divide 3+3\sqrt{17} by -12.

$$x=\frac{-\sqrt{17}-1}{4}$$

Now solve the equation x=\frac{3±3\sqrt{17}}{-12} when ± is minus. Subtract 3\sqrt{17} from 3.

$$x=\frac{3-3\sqrt{17}}{-12}$$

Divide 3-3\sqrt{17} by -12.

$$x=\frac{\sqrt{17}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1-\sqrt{17}}{4} for x_{1} and \frac{-1+\sqrt{17}}{4} for x_{2}.

$$-6x^{2}-3x+6=-6\left(x-\frac{-\sqrt{17}-1}{4}\right)\left(x-\frac{\sqrt{17}-1}{4}\right)$$

Evaluate

$$6-3x-6x^{2}$$