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(h+4)(h+9)=0
We multiply parentheses ..
(+h^2+9h+4h+36)=0
We get rid of parentheses
h^2+9h+4h+36=0
We add all the numbers together, and all the variables
h^2+13h+36=0
a = 1; b = 13; c = +36;
Δ = b2-4ac
Δ = 132-4·1·36
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*1}=\frac{-18}{2} =-9 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*1}=\frac{-8}{2} =-4 $
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