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(x)^2-3x-28=0
a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2} =7 $
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