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16x^2-25x-10=0
a = 16; b = -25; c = -10;
Δ = b2-4ac
Δ = -252-4·16·(-10)
Δ = 1265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{1265}}{2*16}=\frac{25-\sqrt{1265}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{1265}}{2*16}=\frac{25+\sqrt{1265}}{32} $
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