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a+b=-6 ab=-7
To solve the equation, factor t^{2}-6t-7 using formula t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). To find a and b, set up a system to be solved.
a=-7 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(t-7\right)\left(t+1\right)
Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.
t=7 t=-1
To find equation solutions, solve t-7=0 and t+1=0.
a+b=-6 ab=1\left(-7\right)=-7
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-7. To find a and b, set up a system to be solved.
a=-7 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(t^{2}-7t\right)+\left(t-7\right)
Rewrite t^{2}-6t-7 as \left(t^{2}-7t\right)+\left(t-7\right).
t\left(t-7\right)+t-7
Factor out t in t^{2}-7t.
\left(t-7\right)\left(t+1\right)
Factor out common term t-7 by using distributive property.
t=7 t=-1
To find equation solutions, solve t-7=0 and t+1=0.
t^{2}-6t-7=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-7\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-6\right)±\sqrt{36-4\left(-7\right)}}{2}
Square -6.
t=\frac{-\left(-6\right)±\sqrt{36+28}}{2}
Multiply -4 times -7.
t=\frac{-\left(-6\right)±\sqrt{64}}{2}
Add 36 to 28.
t=\frac{-\left(-6\right)±8}{2}
Take the square root of 64.
t=\frac{6±8}{2}
The opposite of -6 is 6.
t=\frac{14}{2}
Now solve the equation t=\frac{6±8}{2} when ± is plus. Add 6 to 8.
t=7
Divide 14 by 2.
t=-\frac{2}{2}
Now solve the equation t=\frac{6±8}{2} when ± is minus. Subtract 8 from 6.
t=-1
Divide -2 by 2.
t=7 t=-1
The equation is now solved.
t^{2}-6t-7=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
t^{2}-6t-7-\left(-7\right)=-\left(-7\right)
Add 7 to both sides of the equation.
t^{2}-6t=-\left(-7\right)
Subtracting -7 from itself leaves 0.
t^{2}-6t=7
Subtract -7 from 0.
t^{2}-6t+\left(-3\right)^{2}=7+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-6t+9=7+9
Square -3.
t^{2}-6t+9=16
Add 7 to 9.
\left(t-3\right)^{2}=16
Factor t^{2}-6t+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(t-3\right)^{2}}=\sqrt{16}
Take the square root of both sides of the equation.
t-3=4 t-3=-4
Simplify.
t=7 t=-1
Add 3 to both sides of the equation.