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3x^2-3x-168=0
a = 3; b = -3; c = -168;
Δ = b2-4ac
Δ = -32-4·3·(-168)
Δ = 2025
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{2025}=45$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-45}{2*3}=\frac{-42}{6} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+45}{2*3}=\frac{48}{6} =8 $
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