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2x^2+17x=210
We move all terms to the left:
2x^2+17x-(210)=0
a = 2; b = 17; c = -210;
Δ = b2-4ac
Δ = 172-4·2·(-210)
Δ = 1969
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{1969}}{2*2}=\frac{-17-\sqrt{1969}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{1969}}{2*2}=\frac{-17+\sqrt{1969}}{4} $
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