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3x^2+28x-96=0
a = 3; b = 28; c = -96;
Δ = b2-4ac
Δ = 282-4·3·(-96)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-44}{2*3}=\frac{-72}{6} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+44}{2*3}=\frac{16}{6} =2+2/3 $
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