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3x^2+45x+16=0
a = 3; b = 45; c = +16;
Δ = b2-4ac
Δ = 452-4·3·16
Δ = 1833
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{-(45)-\sqrt{1833}}{2*3}=\frac{-45-\sqrt{1833}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{1833}}{2*3}=\frac{-45+\sqrt{1833}}{6} $
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