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a+b=6 ab=-72
To solve the equation, factor t^{2}+6t-72 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.
-1,72 -2,36 -3,24 -4,18 -6,12 -8,9
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.
-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1
Calculate the sum for each pair.
a=-6 b=12
The solution is the pair that gives sum 6.
\left(t-6\right)\left(t+12\right)
Rewrite factored expression \left(t+a\right)\left(t+b\right) using the obtained values.
t=6 t=-12
To find equation solutions, solve t-6=0 and t+12=0.
a+b=6 ab=1\left(-72\right)=-72
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as t^{2}+at+bt-72. To find a and b, set up a system to be solved.
-1,72 -2,36 -3,24 -4,18 -6,12 -8,9
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.
-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1
Calculate the sum for each pair.
a=-6 b=12
The solution is the pair that gives sum 6.
\left(t^{2}-6t\right)+\left(12t-72\right)
Rewrite t^{2}+6t-72 as \left(t^{2}-6t\right)+\left(12t-72\right).
t\left(t-6\right)+12\left(t-6\right)
Factor out t in the first and 12 in the second group.
\left(t-6\right)\left(t+12\right)
Factor out common term t-6 by using distributive property.
t=6 t=-12
To find equation solutions, solve t-6=0 and t+12=0.
t^{2}+6t-72=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{-6±\sqrt{6^{2}-4\left(-72\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-6±\sqrt{36-4\left(-72\right)}}{2}
Square 6.
t=\frac{-6±\sqrt{36+288}}{2}
Multiply -4 times -72.
t=\frac{-6±\sqrt{324}}{2}
Add 36 to 288.
t=\frac{-6±18}{2}
Take the square root of 324.
t=\frac{12}{2}
Now solve the equation t=\frac{-6±18}{2} when ± is plus. Add -6 to 18.
t=6
Divide 12 by 2.
t=-\frac{24}{2}
Now solve the equation t=\frac{-6±18}{2} when ± is minus. Subtract 18 from -6.
t=-12
Divide -24 by 2.
t=6 t=-12
The equation is now solved.
t^{2}+6t-72=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-72-\left(-72\right)=-\left(-72\right)
Add 72 to both sides of the equation.
t^{2}+6t=-\left(-72\right)
Subtracting -72 from itself leaves 0.
t^{2}+6t=72
Subtract -72 from 0.
t^{2}+6t+3^{2}=72+3^{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=72+9
Square 3.
t^{2}+6t+9=81
Add 72 to 9.
\left(t+3\right)^{2}=81
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{81}
Take the square root of both sides of the equation.
t+3=9 t+3=-9
Simplify.
t=6 t=-12
Subtract 3 from both sides of the equation.
x ^ 2 +6x -72 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = -6 rs = -72
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -3 - u s = -3 + u
Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -3. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-3 - u) (-3 + u) = -72
To solve for unknown quantity u, substitute these in the product equation rs = -72
9 - u^2 = -72
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -72-9 = -81
Simplify the expression by subtracting 9 on both sides
u^2 = 81 u = \pm\sqrt{81} = \pm 9
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - 9 = -12 s = -3 + 9 = 6
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.