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16x^2+7x-10=0
a = 16; b = 7; c = -10;
Δ = b2-4ac
Δ = 72-4·16·(-10)
Δ = 689
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{-(7)-\sqrt{689}}{2*16}=\frac{-7-\sqrt{689}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{689}}{2*16}=\frac{-7+\sqrt{689}}{32} $
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