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X(3X+7)=49
We move all terms to the left:
X(3X+7)-(49)=0
We multiply parentheses
3X^2+7X-49=0
a = 3; b = 7; c = -49;
Δ = b2-4ac
Δ = 72-4·3·(-49)
Δ = 637
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{637}=\sqrt{49*13}=\sqrt{49}*\sqrt{13}=7\sqrt{13}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7\sqrt{13}}{2*3}=\frac{-7-7\sqrt{13}}{6} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7\sqrt{13}}{2*3}=\frac{-7+7\sqrt{13}}{6} $
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