Solve -3x^2+3x+90

$$-3 { x }^{ 2 } +3x+90$$

Factor

$$-3\left(x-6\right)\left(x+5\right)$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}+x+30\right)$$

Consider -x^{2}+x+30. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=1 ab=-30=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=6 b=-5$$

Rewrite -x^{2}+x+30 as \left(-x^{2}+6x\right)+\left(-5x+30\right).

$$\left(-x^{2}+6x\right)+\left(-5x+30\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(x-6\right)-5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-5\right)$$

Rewrite the complete factored expression.

$$3\left(x-6\right)\left(-x-5\right)$$

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.

$$-3x^{2}+3x+90=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-3\right)\times 90}}{2\left(-3\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-3\right)\times 90}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-3±\sqrt{9+12\times 90}}{2\left(-3\right)}$$

Multiply 12 times 90.

$$x=\frac{-3±\sqrt{9+1080}}{2\left(-3\right)}$$

Add 9 to 1080.

$$x=\frac{-3±\sqrt{1089}}{2\left(-3\right)}$$

Take the square root of 1089.

$$x=\frac{-3±33}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-3±33}{-6}$$

Now solve the equation x=\frac{-3±33}{-6} when ± is plus. Add -3 to 33.

$$x=\frac{30}{-6}$$

Divide 30 by -6.

$$x=-5$$

Now solve the equation x=\frac{-3±33}{-6} when ± is minus. Subtract 33 from -3.

$$x=-\frac{36}{-6}$$

Divide -36 by -6.

$$x=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and 6 for x_{2}.

$$-3x^{2}+3x+90=-3\left(x-\left(-5\right)\right)\left(x-6\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}+3x+90=-3\left(x+5\right)\left(x-6\right)$$

Evaluate

$$-3\left(x-6\right)\left(x+5\right)$$