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2x^2+3x-495=0.
a = 2; b = 3; c = -495;
Δ = b2-4ac
Δ = 32-4·2·(-495)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-63}{2*2}=\frac{-66}{4} =-16+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+63}{2*2}=\frac{60}{4} =15 $
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