Solve 2h^2+28h+90

$$2 h ^ { 2 } + 28 h + 90$$

Factor

$$2\left(h+5\right)\left(h+9\right)$$

Solution Steps

Factor out 2.

$$2\left(h^{2}+14h+45\right)$$

Consider h^{2}+14h+45. Factor the expression by grouping. First, the expression needs to be rewritten as h^{2}+ah+bh+45. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\times 45=45$$

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 45.

$$1,45 3,15 5,9$$

Calculate the sum for each pair.

$$1+45=46 3+15=18 5+9=14$$

The solution is the pair that gives sum 14.

$$a=5 b=9$$

Rewrite h^{2}+14h+45 as \left(h^{2}+5h\right)+\left(9h+45\right).

$$\left(h^{2}+5h\right)+\left(9h+45\right)$$

Factor out h in the first and 9 in the second group.

$$h\left(h+5\right)+9\left(h+5\right)$$

Factor out common term h+5 by using distributive property.

$$\left(h+5\right)\left(h+9\right)$$

Rewrite the complete factored expression.

$$2\left(h+5\right)\left(h+9\right)$$

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.

$$2h^{2}+28h+90=0$$

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.

$$h=\frac{-28±\sqrt{28^{2}-4\times 2\times 90}}{2\times 2}$$

Square 28.

$$h=\frac{-28±\sqrt{784-4\times 2\times 90}}{2\times 2}$$

Multiply -4 times 2.

$$h=\frac{-28±\sqrt{784-8\times 90}}{2\times 2}$$

Multiply -8 times 90.

$$h=\frac{-28±\sqrt{784-720}}{2\times 2}$$

Add 784 to -720.

$$h=\frac{-28±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$h=\frac{-28±8}{2\times 2}$$

Multiply 2 times 2.

$$h=\frac{-28±8}{4}$$

Now solve the equation h=\frac{-28±8}{4} when ± is plus. Add -28 to 8.

$$h=-\frac{20}{4}$$

Divide -20 by 4.

$$h=-5$$

Now solve the equation h=\frac{-28±8}{4} when ± is minus. Subtract 8 from -28.

$$h=-\frac{36}{4}$$

Divide -36 by 4.

$$h=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and -9 for x_{2}.

$$2h^{2}+28h+90=2\left(h-\left(-5\right)\right)\left(h-\left(-9\right)\right)$$

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

$$2h^{2}+28h+90=2\left(h+5\right)\left(h+9\right)$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 +14x +45 = 0$$

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

$$r + s = -14 rs = 45$$

Two numbers r and s sum up to -14 exactly when the average of the two numbers is \frac{1}{2}*-14 = -7. 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>

$$r = -7 - u s = -7 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 45

$$(-7 - u) (-7 + u) = 45$$

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

$$49 - u^2 = 45$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = 45-49 = -4$$

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

$$u^2 = 4 u = \pm\sqrt{4} = \pm 2$$

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

$$r =-7 - 2 = -9 s = -7 + 2 = -5$$

Evaluate

$$2\left(h+5\right)\left(h+9\right)$$