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4x^2+3x=6
We move all terms to the left:
4x^2+3x-(6)=0
a = 4; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·4·(-6)
Δ = 105
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{105}}{2*4}=\frac{-3-\sqrt{105}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{105}}{2*4}=\frac{-3+\sqrt{105}}{8} $
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