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4x^2+300x-5400=0
a = 4; b = 300; c = -5400;
Δ = b2-4ac
Δ = 3002-4·4·(-5400)
Δ = 176400
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{176400}=420$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(300)-420}{2*4}=\frac{-720}{8} =-90 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(300)+420}{2*4}=\frac{120}{8} =15 $
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