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4x^2+16x-1140=0
a = 4; b = 16; c = -1140;
Δ = b2-4ac
Δ = 162-4·4·(-1140)
Δ = 18496
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{18496}=136$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-136}{2*4}=\frac{-152}{8} =-19 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+136}{2*4}=\frac{120}{8} =15 $
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