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X(3X+20)=1463
We move all terms to the left:
X(3X+20)-(1463)=0
We multiply parentheses
3X^2+20X-1463=0
a = 3; b = 20; c = -1463;
Δ = b2-4ac
Δ = 202-4·3·(-1463)
Δ = 17956
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{17956}=134$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-134}{2*3}=\frac{-154}{6} =-25+2/3 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+134}{2*3}=\frac{114}{6} =19 $
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