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4x^2+56x-245=0
a = 4; b = 56; c = -245;
Δ = b2-4ac
Δ = 562-4·4·(-245)
Δ = 7056
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{7056}=84$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-84}{2*4}=\frac{-140}{8} =-17+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+84}{2*4}=\frac{28}{8} =3+1/2 $
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