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5x^2-4x-28=0
a = 5; b = -4; c = -28;
Δ = b2-4ac
Δ = -42-4·5·(-28)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-24}{2*5}=\frac{-20}{10} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+24}{2*5}=\frac{28}{10} =2+4/5 $
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