Solve 20y^2-49y+30

$$20 y ^ { 2 } - 49 y + 30$$

Factor

$$\left(4y-5\right)\left(5y-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20y^{2}+ay+by+30. To find a and b, set up a system to be solved.

$$a+b=-49 ab=20\times 30=600$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 600.

$$-1,-600 -2,-300 -3,-200 -4,-150 -5,-120 -6,-100 -8,-75 -10,-60 -12,-50 -15,-40 -20,-30 -24,-25$$

Calculate the sum for each pair.

$$-1-600=-601 -2-300=-302 -3-200=-203 -4-150=-154 -5-120=-125 -6-100=-106 -8-75=-83 -10-60=-70 -12-50=-62 -15-40=-55 -20-30=-50 -24-25=-49$$

The solution is the pair that gives sum -49.

$$a=-25 b=-24$$

Rewrite 20y^{2}-49y+30 as \left(20y^{2}-25y\right)+\left(-24y+30\right).

$$\left(20y^{2}-25y\right)+\left(-24y+30\right)$$

Factor out 5y in the first and -6 in the second group.

$$5y\left(4y-5\right)-6\left(4y-5\right)$$

Factor out common term 4y-5 by using distributive property.

$$\left(4y-5\right)\left(5y-6\right)$$

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.

$$20y^{2}-49y+30=0$$

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.

$$y=\frac{-\left(-49\right)±\sqrt{\left(-49\right)^{2}-4\times 20\times 30}}{2\times 20}$$

Square -49.

$$y=\frac{-\left(-49\right)±\sqrt{2401-4\times 20\times 30}}{2\times 20}$$

Multiply -4 times 20.

$$y=\frac{-\left(-49\right)±\sqrt{2401-80\times 30}}{2\times 20}$$

Multiply -80 times 30.

$$y=\frac{-\left(-49\right)±\sqrt{2401-2400}}{2\times 20}$$

Add 2401 to -2400.

$$y=\frac{-\left(-49\right)±\sqrt{1}}{2\times 20}$$

Take the square root of 1.

$$y=\frac{-\left(-49\right)±1}{2\times 20}$$

The opposite of -49 is 49.

$$y=\frac{49±1}{2\times 20}$$

Multiply 2 times 20.

$$y=\frac{49±1}{40}$$

Now solve the equation y=\frac{49±1}{40} when ± is plus. Add 49 to 1.

$$y=\frac{50}{40}$$

Reduce the fraction \frac{50}{40} to lowest terms by extracting and canceling out 10.

$$y=\frac{5}{4}$$

Now solve the equation y=\frac{49±1}{40} when ± is minus. Subtract 1 from 49.

$$y=\frac{48}{40}$$

Reduce the fraction \frac{48}{40} to lowest terms by extracting and canceling out 8.

$$y=\frac{6}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and \frac{6}{5} for x_{2}.

$$20y^{2}-49y+30=20\left(y-\frac{5}{4}\right)\left(y-\frac{6}{5}\right)$$

Subtract \frac{5}{4} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}-49y+30=20\times \frac{4y-5}{4}\left(y-\frac{6}{5}\right)$$

Subtract \frac{6}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$20y^{2}-49y+30=20\times \frac{4y-5}{4}\times \frac{5y-6}{5}$$

Multiply \frac{4y-5}{4} times \frac{5y-6}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20y^{2}-49y+30=20\times \frac{\left(4y-5\right)\left(5y-6\right)}{4\times 5}$$

Multiply 4 times 5.

$$20y^{2}-49y+30=20\times \frac{\left(4y-5\right)\left(5y-6\right)}{20}$$

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

$$20y^{2}-49y+30=\left(4y-5\right)\left(5y-6\right)$$

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 20

$$x ^ 2 -\frac{49}{20}x +\frac{3}{2} = 0$$

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

$$r + s = \frac{49}{20} rs = \frac{3}{2}$$

Two numbers r and s sum up to \frac{49}{20} exactly when the average of the two numbers is \frac{1}{2}*\frac{49}{20} = \frac{49}{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>

$$r = \frac{49}{40} - u s = \frac{49}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}

$$(\frac{49}{40} - u) (\frac{49}{40} + u) = \frac{3}{2}$$

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

$$\frac{2401}{1600} - u^2 = \frac{3}{2}$$

Simplify the expression by subtracting \frac{2401}{1600} on both sides

$$-u^2 = \frac{3}{2}-\frac{2401}{1600} = -\frac{1}{1600}$$

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

$$u^2 = \frac{1}{1600} u = \pm\sqrt{\frac{1}{1600}} = \pm \frac{1}{40}$$

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

$$r =\frac{49}{40} - \frac{1}{40} = 1.200 s = \frac{49}{40} + \frac{1}{40} = 1.250$$

Evaluate

$$\left(4y-5\right)\left(5y-6\right)$$