Solve 11x^2+9-4x-8x^2-15x

$$11 x ^ { 2 } + 9 - 4 x - 8 x ^ { 2 } - 15 x$$

Evaluate

$$3x^{2}-19x+9$$

Solution Steps

Combine 11x^{2} and -8x^{2} to get 3x^{2}.

$$3x^{2}+9-4x-15x$$

Combine -4x and -15x to get -19x.

$$3x^{2}+9-19x$$

Factor

$$3\left(x-\frac{19-\sqrt{253}}{6}\right)\left(x-\frac{\sqrt{253}+19}{6}\right)$$

Steps Using the Quadratic Formula

Combine 11x^{2} and -8x^{2} to get 3x^{2}.

$$factor(3x^{2}+9-4x-15x)$$

Combine -4x and -15x to get -19x.

$$factor(3x^{2}+9-19x)$$

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.

$$3x^{2}-19x+9=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 3\times 9}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-19\right)±\sqrt{361-12\times 9}}{2\times 3}$$

Multiply -12 times 9.

$$x=\frac{-\left(-19\right)±\sqrt{361-108}}{2\times 3}$$

Add 361 to -108.

$$x=\frac{-\left(-19\right)±\sqrt{253}}{2\times 3}$$

The opposite of -19 is 19.

$$x=\frac{19±\sqrt{253}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{19±\sqrt{253}}{6}$$

Now solve the equation x=\frac{19±\sqrt{253}}{6} when ± is plus. Add 19 to \sqrt{253}.

$$x=\frac{\sqrt{253}+19}{6}$$

Now solve the equation x=\frac{19±\sqrt{253}}{6} when ± is minus. Subtract \sqrt{253} from 19.

$$x=\frac{19-\sqrt{253}}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{19+\sqrt{253}}{6} for x_{1} and \frac{19-\sqrt{253}}{6} for x_{2}.

$$3x^{2}-19x+9=3\left(x-\frac{\sqrt{253}+19}{6}\right)\left(x-\frac{19-\sqrt{253}}{6}\right)$$