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25x^2+20x+3=0
a = 25; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·25·3
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*25}=\frac{-30}{50} =-3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*25}=\frac{-10}{50} =-1/5 $
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