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3h^2+37h+12=0
a = 3; b = 37; c = +12;
Δ = b2-4ac
Δ = 372-4·3·12
Δ = 1225
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{1225}=35$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(37)-35}{2*3}=\frac{-72}{6} =-12 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+35}{2*3}=\frac{-2}{6} =-1/3 $
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