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4x^2+3x-280=0
a = 4; b = 3; c = -280;
Δ = b2-4ac
Δ = 32-4·4·(-280)
Δ = 4489
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{4489}=67$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-67}{2*4}=\frac{-70}{8} =-8+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+67}{2*4}=\frac{64}{8} =8 $
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