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25x^2+20x=7
We move all terms to the left:
25x^2+20x-(7)=0
a = 25; b = 20; c = -7;
Δ = b2-4ac
Δ = 202-4·25·(-7)
Δ = 1100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1100}=\sqrt{100*11}=\sqrt{100}*\sqrt{11}=10\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{11}}{2*25}=\frac{-20-10\sqrt{11}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{11}}{2*25}=\frac{-20+10\sqrt{11}}{50} $
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