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2h^2+3h+-1224=0
We add all the numbers together, and all the variables
2h^2+3h=0
a = 2; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·2·0
Δ = 9
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{9}=3$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*2}=\frac{-6}{4} =-1+1/2 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*2}=\frac{0}{4} =0 $
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