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4h^2-11h-3=0
a = 4; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·4·(-3)
Δ = 169
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{169}=13$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*4}=\frac{-2}{8} =-1/4 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*4}=\frac{24}{8} =3 $
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