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4x^2-15x+12=0
a = 4; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·4·12
Δ = 33
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{-(-15)-\sqrt{33}}{2*4}=\frac{15-\sqrt{33}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{33}}{2*4}=\frac{15+\sqrt{33}}{8} $
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