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4a^2+3a-11=0
a = 4; b = 3; c = -11;
Δ = b2-4ac
Δ = 32-4·4·(-11)
Δ = 185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{185}}{2*4}=\frac{-3-\sqrt{185}}{8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{185}}{2*4}=\frac{-3+\sqrt{185}}{8} $
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