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10a^2+11a-16=0
a = 10; b = 11; c = -16;
Δ = b2-4ac
Δ = 112-4·10·(-16)
Δ = 761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{761}}{2*10}=\frac{-11-\sqrt{761}}{20} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{761}}{2*10}=\frac{-11+\sqrt{761}}{20} $
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