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4x^2+11x-48=0
a = 4; b = 11; c = -48;
Δ = b2-4ac
Δ = 112-4·4·(-48)
Δ = 889
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{-(11)-\sqrt{889}}{2*4}=\frac{-11-\sqrt{889}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{889}}{2*4}=\frac{-11+\sqrt{889}}{8} $
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