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3x^2+10x=25
We move all terms to the left:
3x^2+10x-(25)=0
a = 3; b = 10; c = -25;
Δ = b2-4ac
Δ = 102-4·3·(-25)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-20}{2*3}=\frac{-30}{6} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+20}{2*3}=\frac{10}{6} =1+2/3 $
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