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7x^2+14x=49
We move all terms to the left:
7x^2+14x-(49)=0
a = 7; b = 14; c = -49;
Δ = b2-4ac
Δ = 142-4·7·(-49)
Δ = 1568
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1568}=\sqrt{784*2}=\sqrt{784}*\sqrt{2}=28\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-28\sqrt{2}}{2*7}=\frac{-14-28\sqrt{2}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+28\sqrt{2}}{2*7}=\frac{-14+28\sqrt{2}}{14} $
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