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2x^2+7x=260
We move all terms to the left:
2x^2+7x-(260)=0
a = 2; b = 7; c = -260;
Δ = b2-4ac
Δ = 72-4·2·(-260)
Δ = 2129
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{-(7)-\sqrt{2129}}{2*2}=\frac{-7-\sqrt{2129}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{2129}}{2*2}=\frac{-7+\sqrt{2129}}{4} $
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