Solve -7y=5y^2-6

$$- 7 y = 5 y ^ { 2 } - 6$$

Solve for y

$$y=-2$$

$$y=\frac{3}{5}=0.6$$

Steps Using Factoring By Grouping

Subtract 5y^{2} from both sides.

$$-7y-5y^{2}=-6$$

Add 6 to both sides.

$$-7y-5y^{2}+6=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5y^{2}-7y+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-5\times 6=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=3 b=-10$$

Rewrite -5y^{2}-7y+6 as \left(-5y^{2}+3y\right)+\left(-10y+6\right).

$$\left(-5y^{2}+3y\right)+\left(-10y+6\right)$$

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

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

Factor out common term 5y-3 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 5y^{2} from both sides.

$$-7y-5y^{2}=-6$$

Add 6 to both sides.

$$-7y-5y^{2}+6=0$$

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.

$$-5y^{2}-7y+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Square -7.

$$y=\frac{-\left(-7\right)±\sqrt{49-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$y=\frac{-\left(-7\right)±\sqrt{49+20\times 6}}{2\left(-5\right)}$$

Multiply 20 times 6.

$$y=\frac{-\left(-7\right)±\sqrt{49+120}}{2\left(-5\right)}$$

Add 49 to 120.

$$y=\frac{-\left(-7\right)±\sqrt{169}}{2\left(-5\right)}$$

Take the square root of 169.

$$y=\frac{-\left(-7\right)±13}{2\left(-5\right)}$$

The opposite of -7 is 7.

$$y=\frac{7±13}{2\left(-5\right)}$$

Multiply 2 times -5.

$$y=\frac{7±13}{-10}$$

Now solve the equation y=\frac{7±13}{-10} when ± is plus. Add 7 to 13.

$$y=\frac{20}{-10}$$

Divide 20 by -10.

$$y=-2$$

Now solve the equation y=\frac{7±13}{-10} when ± is minus. Subtract 13 from 7.

$$y=-\frac{6}{-10}$$

Reduce the fraction \frac{-6}{-10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 5y^{2} from both sides.

$$-7y-5y^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5y^{2}-7y=-6$$

Divide both sides by -5.

$$\frac{-5y^{2}-7y}{-5}=-\frac{6}{-5}$$

Dividing by -5 undoes the multiplication by -5.

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

Divide -7 by -5.

$$y^{2}+\frac{7}{5}y=-\frac{6}{-5}$$

Divide -6 by -5.

$$y^{2}+\frac{7}{5}y=\frac{6}{5}$$

Divide \frac{7}{5}, the coefficient of the x term, by 2 to get \frac{7}{10}. Then add the square of \frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{7}{5}y+\left(\frac{7}{10}\right)^{2}=\frac{6}{5}+\left(\frac{7}{10}\right)^{2}$$

Square \frac{7}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{7}{5}y+\frac{49}{100}=\frac{6}{5}+\frac{49}{100}$$

Add \frac{6}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{7}{5}y+\frac{49}{100}=\frac{169}{100}$$

Factor y^{2}+\frac{7}{5}y+\frac{49}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{7}{10}\right)^{2}=\frac{169}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{7}{10}\right)^{2}}=\sqrt{\frac{169}{100}}$$

Simplify.

$$y+\frac{7}{10}=\frac{13}{10} y+\frac{7}{10}=-\frac{13}{10}$$

Subtract \frac{7}{10} from both sides of the equation.

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