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4x^2+9x-3=0
a = 4; b = 9; c = -3;
Δ = b2-4ac
Δ = 92-4·4·(-3)
Δ = 129
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{-(9)-\sqrt{129}}{2*4}=\frac{-9-\sqrt{129}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{129}}{2*4}=\frac{-9+\sqrt{129}}{8} $
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