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0=3x^2+x-16
We move all terms to the left:
0-(3x^2+x-16)=0
We add all the numbers together, and all the variables
-(3x^2+x-16)=0
We get rid of parentheses
-3x^2-x+16=0
We add all the numbers together, and all the variables
-3x^2-1x+16=0
a = -3; b = -1; c = +16;
Δ = b2-4ac
Δ = -12-4·(-3)·16
Δ = 193
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{-(-1)-\sqrt{193}}{2*-3}=\frac{1-\sqrt{193}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{193}}{2*-3}=\frac{1+\sqrt{193}}{-6} $
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