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4x^2-9x-28=0
a = 4; b = -9; c = -28;
Δ = b2-4ac
Δ = -92-4·4·(-28)
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-23}{2*4}=\frac{-14}{8} =-1+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+23}{2*4}=\frac{32}{8} =4 $
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