Solve 7y+y^2-30

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

Factor

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

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 positive, the positive number has greater absolute value than the negative. 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 y^{2}+7y-30 as \left(y^{2}-3y\right)+\left(10y-30\right).

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

Factor out y in the first and 10 in the second group.

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

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

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

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.

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

Square 7.

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

Multiply -4 times -30.

$$y=\frac{-7±\sqrt{49+120}}{2}$$

Add 49 to 120.

$$y=\frac{-7±\sqrt{169}}{2}$$

Take the square root of 169.

$$y=\frac{-7±13}{2}$$

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

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

Divide 6 by 2.

$$y=3$$

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

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

Divide -20 by 2.

$$y=-10$$

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

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

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

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

Evaluate

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