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3x^2+2x-61=-20
We move all terms to the left:
3x^2+2x-61-(-20)=0
We add all the numbers together, and all the variables
3x^2+2x-41=0
a = 3; b = 2; c = -41;
Δ = b2-4ac
Δ = 22-4·3·(-41)
Δ = 496
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{31}}{2*3}=\frac{-2-4\sqrt{31}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{31}}{2*3}=\frac{-2+4\sqrt{31}}{6} $
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