Solve 20x^2-30x-200

$$20 x ^ { 2 } - 30 x - 200$$

Factor

$$10\left(x-4\right)\left(2x+5\right)$$

Solution Steps

Factor out 10.

$$10\left(2x^{2}-3x-20\right)$$

Consider 2x^{2}-3x-20. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-20\right)=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

Rewrite 2x^{2}-3x-20 as \left(2x^{2}-8x\right)+\left(5x-20\right).

$$\left(2x^{2}-8x\right)+\left(5x-20\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(x-4\right)+5\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(2x+5\right)$$

Rewrite the complete factored expression.

$$10\left(x-4\right)\left(2x+5\right)$$

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.

$$20x^{2}-30x-200=0$$

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.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 20\left(-200\right)}}{2\times 20}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 20\left(-200\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-30\right)±\sqrt{900-80\left(-200\right)}}{2\times 20}$$

Multiply -80 times -200.

$$x=\frac{-\left(-30\right)±\sqrt{900+16000}}{2\times 20}$$

Add 900 to 16000.

$$x=\frac{-\left(-30\right)±\sqrt{16900}}{2\times 20}$$

Take the square root of 16900.

$$x=\frac{-\left(-30\right)±130}{2\times 20}$$

The opposite of -30 is 30.

$$x=\frac{30±130}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{30±130}{40}$$

Now solve the equation x=\frac{30±130}{40} when ± is plus. Add 30 to 130.

$$x=\frac{160}{40}$$

Divide 160 by 40.

$$x=4$$

Now solve the equation x=\frac{30±130}{40} when ± is minus. Subtract 130 from 30.

$$x=-\frac{100}{40}$$

Reduce the fraction \frac{-100}{40} to lowest terms by extracting and canceling out 20.

$$x=-\frac{5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -\frac{5}{2} for x_{2}.

$$20x^{2}-30x-200=20\left(x-4\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$20x^{2}-30x-200=20\left(x-4\right)\left(x+\frac{5}{2}\right)$$

Add \frac{5}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}-30x-200=20\left(x-4\right)\times \frac{2x+5}{2}$$

Cancel out 2, the greatest common factor in 20 and 2.

$$20x^{2}-30x-200=10\left(x-4\right)\left(2x+5\right)$$

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 20

$$x ^ 2 -\frac{3}{2}x -10 = 0$$

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

$$r + s = \frac{3}{2} rs = -10$$

Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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>

$$r = \frac{3}{4} - u s = \frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -10

$$(\frac{3}{4} - u) (\frac{3}{4} + u) = -10$$

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

$$\frac{9}{16} - u^2 = -10$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -10-\frac{9}{16} = -\frac{169}{16}$$

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

$$u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{4}$$

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

$$r =\frac{3}{4} - \frac{13}{4} = -2.500 s = \frac{3}{4} + \frac{13}{4} = 4$$

Evaluate

$$10\left(x-4\right)\left(2x+5\right)$$