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4x^2-25x-86=0
a = 4; b = -25; c = -86;
Δ = b2-4ac
Δ = -252-4·4·(-86)
Δ = 2001
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{2001}}{2*4}=\frac{25-\sqrt{2001}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{2001}}{2*4}=\frac{25+\sqrt{2001}}{8} $
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