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25=3x^2-10x
We move all terms to the left:
25-(3x^2-10x)=0
We get rid of parentheses
-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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