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2x^2+15x=27
We move all terms to the left:
2x^2+15x-(27)=0
a = 2; b = 15; c = -27;
Δ = b2-4ac
Δ = 152-4·2·(-27)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-21}{2*2}=\frac{-36}{4} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+21}{2*2}=\frac{6}{4} =1+1/2 $
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