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25x^2+16x=80
We move all terms to the left:
25x^2+16x-(80)=0
a = 25; b = 16; c = -80;
Δ = b2-4ac
Δ = 162-4·25·(-80)
Δ = 8256
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8256}=\sqrt{64*129}=\sqrt{64}*\sqrt{129}=8\sqrt{129}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{129}}{2*25}=\frac{-16-8\sqrt{129}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{129}}{2*25}=\frac{-16+8\sqrt{129}}{50} $
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