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16x^2+36x+10=0
a = 16; b = 36; c = +10;
Δ = b2-4ac
Δ = 362-4·16·10
Δ = 656
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{41}}{2*16}=\frac{-36-4\sqrt{41}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{41}}{2*16}=\frac{-36+4\sqrt{41}}{32} $
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