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15x+4x^2-3=0
a = 4; b = 15; c = -3;
Δ = b2-4ac
Δ = 152-4·4·(-3)
Δ = 273
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{273}}{2*4}=\frac{-15-\sqrt{273}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{273}}{2*4}=\frac{-15+\sqrt{273}}{8} $
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