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3x^2+2x-5-14=0
We add all the numbers together, and all the variables
3x^2+2x-19=0
a = 3; b = 2; c = -19;
Δ = b2-4ac
Δ = 22-4·3·(-19)
Δ = 232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{58}}{2*3}=\frac{-2-2\sqrt{58}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{58}}{2*3}=\frac{-2+2\sqrt{58}}{6} $
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