Solve y^2+y-90

$$y ^ { 2 } + y - 90$$

Factor

$$\left(y-9\right)\left(y+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-90. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-90\right)=-90$$

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 -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 1.

$$a=-9 b=10$$

Rewrite y^{2}+y-90 as \left(y^{2}-9y\right)+\left(10y-90\right).

$$\left(y^{2}-9y\right)+\left(10y-90\right)$$

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

$$y\left(y-9\right)+10\left(y-9\right)$$

Factor out common term y-9 by using distributive property.

$$\left(y-9\right)\left(y+10\right)$$

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.

$$y^{2}+y-90=0$$

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.

$$y=\frac{-1±\sqrt{1^{2}-4\left(-90\right)}}{2}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\left(-90\right)}}{2}$$

Multiply -4 times -90.

$$y=\frac{-1±\sqrt{1+360}}{2}$$

Add 1 to 360.

$$y=\frac{-1±\sqrt{361}}{2}$$

Take the square root of 361.

$$y=\frac{-1±19}{2}$$

Now solve the equation y=\frac{-1±19}{2} when ± is plus. Add -1 to 19.

$$y=\frac{18}{2}$$

Divide 18 by 2.

$$y=9$$

Now solve the equation y=\frac{-1±19}{2} when ± is minus. Subtract 19 from -1.

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

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 9 for x_{1} and -10 for x_{2}.

$$y^{2}+y-90=\left(y-9\right)\left(y-\left(-10\right)\right)$$

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

$$y^{2}+y-90=\left(y-9\right)\left(y+10\right)$$

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.

$$x ^ 2 +1x -90 = 0$$

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

$$r + s = -1 rs = -90$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -90

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -90$$

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

$$\frac{1}{4} - u^2 = -90$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -90-\frac{1}{4} = -\frac{361}{4}$$

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

$$u^2 = \frac{361}{4} u = \pm\sqrt{\frac{361}{4}} = \pm \frac{19}{2}$$

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

$$r =-\frac{1}{2} - \frac{19}{2} = -10 s = -\frac{1}{2} + \frac{19}{2} = 9$$

Evaluate

$$\left(y-9\right)\left(y+10\right)$$