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9x^2+42x=49
We move all terms to the left:
9x^2+42x-(49)=0
a = 9; b = 42; c = -49;
Δ = b2-4ac
Δ = 422-4·9·(-49)
Δ = 3528
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42\sqrt{2}}{2*9}=\frac{-42-42\sqrt{2}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42\sqrt{2}}{2*9}=\frac{-42+42\sqrt{2}}{18} $
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