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2x^2+28x-190=0
a = 2; b = 28; c = -190;
Δ = b2-4ac
Δ = 282-4·2·(-190)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-48}{2*2}=\frac{-76}{4} =-19 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+48}{2*2}=\frac{20}{4} =5 $
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