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25x^2-30x-5=0
a = 25; b = -30; c = -5;
Δ = b2-4ac
Δ = -302-4·25·(-5)
Δ = 1400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1400}=\sqrt{100*14}=\sqrt{100}*\sqrt{14}=10\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10\sqrt{14}}{2*25}=\frac{30-10\sqrt{14}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10\sqrt{14}}{2*25}=\frac{30+10\sqrt{14}}{50} $
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