Solve 6y^2-7y-20=

$$6 y ^ { 2 } - 7 y - 20 =$$

Factor

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

Solution Steps

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

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

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 -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.

$$a=-15 b=8$$

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

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

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

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

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

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

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.

$$6y^{2}-7y-20=0$$

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.

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

Square -7.

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

Multiply -4 times 6.

$$y=\frac{-\left(-7\right)±\sqrt{49-24\left(-20\right)}}{2\times 6}$$

Multiply -24 times -20.

$$y=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 6}$$

Add 49 to 480.

$$y=\frac{-\left(-7\right)±\sqrt{529}}{2\times 6}$$

Take the square root of 529.

$$y=\frac{-\left(-7\right)±23}{2\times 6}$$

The opposite of -7 is 7.

$$y=\frac{7±23}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{7±23}{12}$$

Now solve the equation y=\frac{7±23}{12} when ± is plus. Add 7 to 23.

$$y=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation y=\frac{7±23}{12} when ± is minus. Subtract 23 from 7.

$$y=-\frac{16}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

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

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

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

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

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

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

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

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

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

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

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

Multiply 2 times 3.

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

Cancel out 6, the greatest common factor in 6 and 6.

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

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 6

$$x ^ 2 -\frac{7}{6}x -\frac{10}{3} = 0$$

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

$$r + s = \frac{7}{6} rs = -\frac{10}{3}$$

Two numbers r and s sum up to \frac{7}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{6} = \frac{7}{12}. 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>

$$r = \frac{7}{12} - u s = \frac{7}{12} + u$$

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

$$(\frac{7}{12} - u) (\frac{7}{12} + u) = -\frac{10}{3}$$

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

$$\frac{49}{144} - u^2 = -\frac{10}{3}$$

Simplify the expression by subtracting \frac{49}{144} on both sides

$$-u^2 = -\frac{10}{3}-\frac{49}{144} = -\frac{529}{144}$$

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

$$u^2 = \frac{529}{144} u = \pm\sqrt{\frac{529}{144}} = \pm \frac{23}{12}$$

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

$$r =\frac{7}{12} - \frac{23}{12} = -1.333 s = \frac{7}{12} + \frac{23}{12} = 2.500$$

Evaluate

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