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2x^2+17x=21
We move all terms to the left:
2x^2+17x-(21)=0
a = 2; b = 17; c = -21;
Δ = b2-4ac
Δ = 172-4·2·(-21)
Δ = 457
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{457}}{2*2}=\frac{-17-\sqrt{457}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{457}}{2*2}=\frac{-17+\sqrt{457}}{4} $
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