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2x^2+7x-5280=0
a = 2; b = 7; c = -5280;
Δ = b2-4ac
Δ = 72-4·2·(-5280)
Δ = 42289
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{42289}}{2*2}=\frac{-7-\sqrt{42289}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{42289}}{2*2}=\frac{-7+\sqrt{42289}}{4} $
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