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2x^2-4x-390=0
a = 2; b = -4; c = -390;
Δ = b2-4ac
Δ = -42-4·2·(-390)
Δ = 3136
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{3136}=56$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-56}{2*2}=\frac{-52}{4} =-13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+56}{2*2}=\frac{60}{4} =15 $
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