Solve -7y+y^2-30

$$-7y+ { y }^{ 2 } -30$$

Factor

$$\left(y-10\right)\left(y+3\right)$$

Solution Steps

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

$$y^{2}-7y-30$$

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

$$a+b=-7 ab=1\left(-30\right)=-30$$

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.

$$a=-10 b=3$$

Rewrite y^{2}-7y-30 as \left(y^{2}-10y\right)+\left(3y-30\right).

$$\left(y^{2}-10y\right)+\left(3y-30\right)$$

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

$$y\left(y-10\right)+3\left(y-10\right)$$

Factor out common term y-10 by using distributive property.

$$\left(y-10\right)\left(y+3\right)$$

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.

$$y^{2}-7y-30=0$$

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.

$$y=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-30\right)}}{2}$$

Square -7.

$$y=\frac{-\left(-7\right)±\sqrt{49-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

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

Add 49 to 120.

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

Take the square root of 169.

$$y=\frac{-\left(-7\right)±13}{2}$$

The opposite of -7 is 7.

$$y=\frac{7±13}{2}$$

Now solve the equation y=\frac{7±13}{2} when ± is plus. Add 7 to 13.

$$y=\frac{20}{2}$$

Divide 20 by 2.

$$y=10$$

Now solve the equation y=\frac{7±13}{2} when ± is minus. Subtract 13 from 7.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -3 for x_{2}.

$$y^{2}-7y-30=\left(y-10\right)\left(y-\left(-3\right)\right)$$

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

$$y^{2}-7y-30=\left(y-10\right)\left(y+3\right)$$

Evaluate

$$\left(y-10\right)\left(y+3\right)$$