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25x^2+40x=20
We move all terms to the left:
25x^2+40x-(20)=0
a = 25; b = 40; c = -20;
Δ = b2-4ac
Δ = 402-4·25·(-20)
Δ = 3600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-60}{2*25}=\frac{-100}{50} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+60}{2*25}=\frac{20}{50} =2/5 $
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