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4x^2-28x-196=0
a = 4; b = -28; c = -196;
Δ = b2-4ac
Δ = -282-4·4·(-196)
Δ = 3920
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3920}=\sqrt{784*5}=\sqrt{784}*\sqrt{5}=28\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28\sqrt{5}}{2*4}=\frac{28-28\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28\sqrt{5}}{2*4}=\frac{28+28\sqrt{5}}{8} $
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