Question 238110
t^2 - 17 = 2t


subtract 2t from both sides of the equation to get:


t^2 - 2t - 17 = 0


that's a quadratic equation.


it doesn't look like it will factor, so use the quadratic formula to solve it.


the quadratic formula is:


x = -b +/- sqrt(b^2-4ac)/2a


standard form of the quadratic equation is ax^2 + bx + c


in your equation of t^2 - 2t - 17 = 0, replace t with x to make it:


x^2 - 2x - 17 = 0


a = coefficient of x^2 term = 1
b = coefficient of x term = -2
c = constant term = -17


substitute in quadratic formula of:


x = -b +/- sqrt(b^2-4ac)/2a to get:


{{{x = (-(-2) +- sqrt((-2)^2 - (4*1*(-17)))) / 2}}}


this becomes


{{{x = (2 +- sqrt(72)) / 2}}}


x = 5.242640687 or x = -3.242640687


to confirm these values are true, substitute in the original equation and solve.


both values substituted and confirmed the original equations so the answers are good.


a graph of this equation looks like:


{{{graph (400,400,-10,10,-20,20,x^2 -2x - 17)}}}