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4x^2+x-1.15=0
a = 4; b = 1; c = -1.15;
Δ = b2-4ac
Δ = 12-4·4·(-1.15)
Δ = 19.4
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{-(1)-\sqrt{19.4}}{2*4}=\frac{-1-\sqrt{19.4}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{19.4}}{2*4}=\frac{-1+\sqrt{19.4}}{8} $
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