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2h^2+5h=12
We move all terms to the left:
2h^2+5h-(12)=0
a = 2; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·2·(-12)
Δ = 121
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{121}=11$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*2}=\frac{-16}{4} =-4 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*2}=\frac{6}{4} =1+1/2 $
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