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4x^2+33x-150=0
a = 4; b = 33; c = -150;
Δ = b2-4ac
Δ = 332-4·4·(-150)
Δ = 3489
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{-(33)-\sqrt{3489}}{2*4}=\frac{-33-\sqrt{3489}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{3489}}{2*4}=\frac{-33+\sqrt{3489}}{8} $
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