Solve 20x^2+7x-6

$$20 x ^ { 2 } + 7 x - 6$$

Factor

$$\left(5x-2\right)\left(4x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=7 ab=20\left(-6\right)=-120$$

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 -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 7.

$$a=-8 b=15$$

Rewrite 20x^{2}+7x-6 as \left(20x^{2}-8x\right)+\left(15x-6\right).

$$\left(20x^{2}-8x\right)+\left(15x-6\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(5x-2\right)+3\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(4x+3\right)$$

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.

$$20x^{2}+7x-6=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 20\left(-6\right)}}{2\times 20}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 20\left(-6\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-7±\sqrt{49-80\left(-6\right)}}{2\times 20}$$

Multiply -80 times -6.

$$x=\frac{-7±\sqrt{49+480}}{2\times 20}$$

Add 49 to 480.

$$x=\frac{-7±\sqrt{529}}{2\times 20}$$

Take the square root of 529.

$$x=\frac{-7±23}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-7±23}{40}$$

Now solve the equation x=\frac{-7±23}{40} when ± is plus. Add -7 to 23.

$$x=\frac{16}{40}$$

Reduce the fraction \frac{16}{40} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-7±23}{40} when ± is minus. Subtract 23 from -7.

$$x=-\frac{30}{40}$$

Reduce the fraction \frac{-30}{40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and -\frac{3}{4} for x_{2}.

$$20x^{2}+7x-6=20\left(x-\frac{2}{5}\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$20x^{2}+7x-6=20\left(x-\frac{2}{5}\right)\left(x+\frac{3}{4}\right)$$

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

$$20x^{2}+7x-6=20\times \frac{5x-2}{5}\left(x+\frac{3}{4}\right)$$

Add \frac{3}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+7x-6=20\times \frac{5x-2}{5}\times \frac{4x+3}{4}$$

Multiply \frac{5x-2}{5} times \frac{4x+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+7x-6=20\times \frac{\left(5x-2\right)\left(4x+3\right)}{5\times 4}$$

Multiply 5 times 4.

$$20x^{2}+7x-6=20\times \frac{\left(5x-2\right)\left(4x+3\right)}{20}$$

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

$$20x^{2}+7x-6=\left(5x-2\right)\left(4x+3\right)$$

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

$$x ^ 2 +\frac{7}{20}x -\frac{3}{10} = 0$$

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

$$r + s = -\frac{7}{20} rs = -\frac{3}{10}$$

Two numbers r and s sum up to -\frac{7}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{20} = -\frac{7}{40}. 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>

$$r = -\frac{7}{40} - u s = -\frac{7}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}

$$(-\frac{7}{40} - u) (-\frac{7}{40} + u) = -\frac{3}{10}$$

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

$$\frac{49}{1600} - u^2 = -\frac{3}{10}$$

Simplify the expression by subtracting \frac{49}{1600} on both sides

$$-u^2 = -\frac{3}{10}-\frac{49}{1600} = -\frac{529}{1600}$$

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

$$u^2 = \frac{529}{1600} u = \pm\sqrt{\frac{529}{1600}} = \pm \frac{23}{40}$$

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

$$r =-\frac{7}{40} - \frac{23}{40} = -0.750 s = -\frac{7}{40} + \frac{23}{40} = 0.400$$

Evaluate

$$\left(5x-2\right)\left(4x+3\right)$$