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2x^2+7x-490=0
a = 2; b = 7; c = -490;
Δ = b2-4ac
Δ = 72-4·2·(-490)
Δ = 3969
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{3969}=63$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-63}{2*2}=\frac{-70}{4} =-17+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+63}{2*2}=\frac{56}{4} =14 $
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