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-16x^2-25x+305=0
a = -16; b = -25; c = +305;
Δ = b2-4ac
Δ = -252-4·(-16)·305
Δ = 20145
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{20145}}{2*-16}=\frac{25-\sqrt{20145}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{20145}}{2*-16}=\frac{25+\sqrt{20145}}{-32} $
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