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4x^2+200x-2600=0
a = 4; b = 200; c = -2600;
Δ = b2-4ac
Δ = 2002-4·4·(-2600)
Δ = 81600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{81600}=\sqrt{1600*51}=\sqrt{1600}*\sqrt{51}=40\sqrt{51}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-40\sqrt{51}}{2*4}=\frac{-200-40\sqrt{51}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+40\sqrt{51}}{2*4}=\frac{-200+40\sqrt{51}}{8} $
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