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2h^2+11h+12=0
a = 2; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·2·12
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$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*2}=\frac{-16}{4} =-4 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*2}=\frac{-6}{4} =-1+1/2 $
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