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3x^2-43x-98=0
a = 3; b = -43; c = -98;
Δ = b2-4ac
Δ = -432-4·3·(-98)
Δ = 3025
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{3025}=55$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-55}{2*3}=\frac{-12}{6} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+55}{2*3}=\frac{98}{6} =16+1/3 $
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