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3x^2+2x+14=19
We move all terms to the left:
3x^2+2x+14-(19)=0
We add all the numbers together, and all the variables
3x^2+2x-5=0
a = 3; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·3·(-5)
Δ = 64
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*3}=\frac{-10}{6} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*3}=\frac{6}{6} =1 $
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