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2x^2+2x-420=0
a = 2; b = 2; c = -420;
Δ = b2-4ac
Δ = 22-4·2·(-420)
Δ = 3364
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{3364}=58$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-58}{2*2}=\frac{-60}{4} =-15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+58}{2*2}=\frac{56}{4} =14 $
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