Let the equation be x^2 - 4x - 14 = -2x2−4x−14=−2 so that the variables in our equation don't conflict with the parameters in the quadratic formula.
The quadratic formula is x = (-b +- sqrt(b^2 - 4ac))/(2a)x=−b±√b2−4ac2a, for equations of the form ax^2 + bx + c = 0ax2+bx+c=0.
Sending all our terms to one side:
x^2 - 4x - 14 = -2x2−4x−14=−2
x^2 - 4x - 14 + 2 = 0x2−4x−14+2=0
x^2 - 4x - 12 = 0x2−4x−12=0
Applying the formula:
x = (-(-4) +- sqrt((-4)^2 - (4 xx 1 xx -12)))/(2 xx 1)x=−(−4)±√(−4)2−(4×1×−12)2×1
x = (4 +- sqrt(16+ 48))/2x=4±√16+482
x = (4 +- sqrt(64))/2x=4±√642
x = (4 +- 8) /2x=4±82
x = (4 + 8)/2 and (4 - 8)/2x=4+82and4−82
x = 12/2 and -4/2x=122and−42
x = 6 and -2x=6and−2
Checking in the original equation , we have that both solutions work!
Therefore, the solution set is {x = 6" & "-2}{x=6 & −2}.
Hopefully this helps!
