Solve 6=-20y^2+23y

$$6 = - 20 y ^ { 2 } + 23 y$$

Solve for y

$$y=\frac{2}{5}=0.4$$

$$y=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$-20y^{2}+23y=6$$

Subtract 6 from both sides.

$$-20y^{2}+23y-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -20y^{2}+ay+by-6. To find a and b, set up a system to be solved.

$$a+b=23 ab=-20\left(-6\right)=120$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both 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.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 23.

$$a=15 b=8$$

Rewrite -20y^{2}+23y-6 as \left(-20y^{2}+15y\right)+\left(8y-6\right).

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

Factor out -5y in the first and 2 in the second group.

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

Factor out common term 4y-3 by using distributive property.

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

To find equation solutions, solve 4y-3=0 and -5y+2=0.

$$y=\frac{3}{4} y=\frac{2}{5}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-20y^{2}+23y=6$$

Subtract 6 from both sides.

$$-20y^{2}+23y-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 23 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-23±\sqrt{23^{2}-4\left(-20\right)\left(-6\right)}}{2\left(-20\right)}$$

Square 23.

$$y=\frac{-23±\sqrt{529-4\left(-20\right)\left(-6\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$y=\frac{-23±\sqrt{529+80\left(-6\right)}}{2\left(-20\right)}$$

Multiply 80 times -6.

$$y=\frac{-23±\sqrt{529-480}}{2\left(-20\right)}$$

Add 529 to -480.

$$y=\frac{-23±\sqrt{49}}{2\left(-20\right)}$$

Take the square root of 49.

$$y=\frac{-23±7}{2\left(-20\right)}$$

Multiply 2 times -20.

$$y=\frac{-23±7}{-40}$$

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

$$y=-\frac{16}{-40}$$

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

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

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

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

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

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

The equation is now solved.

$$y=\frac{2}{5} y=\frac{3}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-20y^{2}+23y=6$$

Divide both sides by -20.

$$\frac{-20y^{2}+23y}{-20}=\frac{6}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$y^{2}+\frac{23}{-20}y=\frac{6}{-20}$$

Divide 23 by -20.

$$y^{2}-\frac{23}{20}y=\frac{6}{-20}$$

Reduce the fraction \frac{6}{-20} to lowest terms by extracting and canceling out 2.

$$y^{2}-\frac{23}{20}y=-\frac{3}{10}$$

Divide -\frac{23}{20}, the coefficient of the x term, by 2 to get -\frac{23}{40}. Then add the square of -\frac{23}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{23}{20}y+\left(-\frac{23}{40}\right)^{2}=-\frac{3}{10}+\left(-\frac{23}{40}\right)^{2}$$

Square -\frac{23}{40} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{23}{20}y+\frac{529}{1600}=-\frac{3}{10}+\frac{529}{1600}$$

Add -\frac{3}{10} to \frac{529}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{23}{20}y+\frac{529}{1600}=\frac{49}{1600}$$

Factor y^{2}-\frac{23}{20}y+\frac{529}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{23}{40}\right)^{2}=\frac{49}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{23}{40}\right)^{2}}=\sqrt{\frac{49}{1600}}$$

Simplify.

$$y-\frac{23}{40}=\frac{7}{40} y-\frac{23}{40}=-\frac{7}{40}$$

Add \frac{23}{40} to both sides of the equation.

$$y=\frac{3}{4} y=\frac{2}{5}$$