Solve 5x^2+20x-60=

$$5 x ^ { 2 } + 20 x - 60 =$$

Factor

$$5\left(x-2\right)\left(x+6\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}+4x-12\right)$$

Consider x^{2}+4x-12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite x^{2}+4x-12 as \left(x^{2}-2x\right)+\left(6x-12\right).

$$\left(x^{2}-2x\right)+\left(6x-12\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-2\right)+6\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$5\left(x-2\right)\left(x+6\right)$$

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.

$$5x^{2}+20x-60=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 5\left(-60\right)}}{2\times 5}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 5\left(-60\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-20±\sqrt{400-20\left(-60\right)}}{2\times 5}$$

Multiply -20 times -60.

$$x=\frac{-20±\sqrt{400+1200}}{2\times 5}$$

Add 400 to 1200.

$$x=\frac{-20±\sqrt{1600}}{2\times 5}$$

Take the square root of 1600.

$$x=\frac{-20±40}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-20±40}{10}$$

Now solve the equation x=\frac{-20±40}{10} when ± is plus. Add -20 to 40.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{-20±40}{10} when ± is minus. Subtract 40 from -20.

$$x=-\frac{60}{10}$$

Divide -60 by 10.

$$x=-6$$

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

$$5x^{2}+20x-60=5\left(x-2\right)\left(x-\left(-6\right)\right)$$

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

$$5x^{2}+20x-60=5\left(x-2\right)\left(x+6\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 5

$$x ^ 2 +4x -12 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -4 rs = -12$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -12

$$(-2 - u) (-2 + u) = -12$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$4 - u^2 = -12$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -12-4 = -16$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 16 u = \pm\sqrt{16} = \pm 4$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-2 - 4 = -6 s = -2 + 4 = 2$$

Evaluate

$$5\left(x-2\right)\left(x+6\right)$$