Solve -3quad2x^2+9x+10

$$- 3 \quad 2 x ^ { 2 } + 9 x + 10$$

Factor

$$-6\left(x-\left(-\frac{\sqrt{321}}{12}+\frac{3}{4}\right)\right)\left(x-\left(\frac{\sqrt{321}}{12}+\frac{3}{4}\right)\right)$$

Steps Using the Quadratic Formula

Multiply -3 and 2 to get -6.

$$factor(-6x^{2}+9x+10)$$

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.

$$-6x^{2}+9x+10=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-6\right)\times 10}}{2\left(-6\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-6\right)\times 10}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-9±\sqrt{81+24\times 10}}{2\left(-6\right)}$$

Multiply 24 times 10.

$$x=\frac{-9±\sqrt{81+240}}{2\left(-6\right)}$$

Add 81 to 240.

$$x=\frac{-9±\sqrt{321}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-9±\sqrt{321}}{-12}$$

Now solve the equation x=\frac{-9±\sqrt{321}}{-12} when ± is plus. Add -9 to \sqrt{321}.

$$x=\frac{\sqrt{321}-9}{-12}$$

Divide -9+\sqrt{321} by -12.

$$x=-\frac{\sqrt{321}}{12}+\frac{3}{4}$$

Now solve the equation x=\frac{-9±\sqrt{321}}{-12} when ± is minus. Subtract \sqrt{321} from -9.

$$x=\frac{-\sqrt{321}-9}{-12}$$

Divide -9-\sqrt{321} by -12.

$$x=\frac{\sqrt{321}}{12}+\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{4}-\frac{\sqrt{321}}{12} for x_{1} and \frac{3}{4}+\frac{\sqrt{321}}{12} for x_{2}.

$$-6x^{2}+9x+10=-6\left(x-\left(-\frac{\sqrt{321}}{12}+\frac{3}{4}\right)\right)\left(x-\left(\frac{\sqrt{321}}{12}+\frac{3}{4}\right)\right)$$

Evaluate

$$10+9x-6x^{2}$$

Solution Steps

Multiply -3 and 2 to get -6.

$$-6x^{2}+9x+10$$