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50x+12=-25x^2
We move all terms to the left:
50x+12-(-25x^2)=0
We get rid of parentheses
25x^2+50x+12=0
a = 25; b = 50; c = +12;
Δ = b2-4ac
Δ = 502-4·25·12
Δ = 1300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{13}}{2*25}=\frac{-50-10\sqrt{13}}{50} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{13}}{2*25}=\frac{-50+10\sqrt{13}}{50} $
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