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2x^2+27x-405=0
a = 2; b = 27; c = -405;
Δ = b2-4ac
Δ = 272-4·2·(-405)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-63}{2*2}=\frac{-90}{4} =-22+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+63}{2*2}=\frac{36}{4} =9 $
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