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16x^2+25x-356=0
a = 16; b = 25; c = -356;
Δ = b2-4ac
Δ = 252-4·16·(-356)
Δ = 23409
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{23409}=153$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-153}{2*16}=\frac{-178}{32} =-5+9/16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+153}{2*16}=\frac{128}{32} =4 $
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