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4x^2-3x-95=0
a = 4; b = -3; c = -95;
Δ = b2-4ac
Δ = -32-4·4·(-95)
Δ = 1529
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{-(-3)-\sqrt{1529}}{2*4}=\frac{3-\sqrt{1529}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1529}}{2*4}=\frac{3+\sqrt{1529}}{8} $
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