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10(3x^2-13x)=-40
We move all terms to the left:
10(3x^2-13x)-(-40)=0
We add all the numbers together, and all the variables
10(3x^2-13x)+40=0
We multiply parentheses
30x^2-130x+40=0
a = 30; b = -130; c = +40;
Δ = b2-4ac
Δ = -1302-4·30·40
Δ = 12100
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{12100}=110$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-130)-110}{2*30}=\frac{20}{60} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-130)+110}{2*30}=\frac{240}{60} =4 $
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