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25x^2+1=20x
We move all terms to the left:
25x^2+1-(20x)=0
a = 25; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·25·1
Δ = 300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{3}}{2*25}=\frac{20-10\sqrt{3}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{3}}{2*25}=\frac{20+10\sqrt{3}}{50} $
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