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h^2+25h-6=0
a = 1; b = 25; c = -6;
Δ = b2-4ac
Δ = 252-4·1·(-6)
Δ = 649
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{649}}{2*1}=\frac{-25-\sqrt{649}}{2} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{649}}{2*1}=\frac{-25+\sqrt{649}}{2} $
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