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a+b=-63 ab=50\left(-18\right)=-900
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 50t^{2}+at+bt-18. To find a and b, set up a system to be solved.
1,-900 2,-450 3,-300 4,-225 5,-180 6,-150 9,-100 10,-90 12,-75 15,-60 18,-50 20,-45 25,-36 30,-30
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. List all such integer pairs that give product -900.
1-900=-899 2-450=-448 3-300=-297 4-225=-221 5-180=-175 6-150=-144 9-100=-91 10-90=-80 12-75=-63 15-60=-45 18-50=-32 20-45=-25 25-36=-11 30-30=0
Calculate the sum for each pair.
a=-75 b=12
The solution is the pair that gives sum -63.
\left(50t^{2}-75t\right)+\left(12t-18\right)
Rewrite 50t^{2}-63t-18 as \left(50t^{2}-75t\right)+\left(12t-18\right).
25t\left(2t-3\right)+6\left(2t-3\right)
Factor out 25t in the first and 6 in the second group.
\left(2t-3\right)\left(25t+6\right)
Factor out common term 2t-3 by using distributive property.
t=\frac{3}{2} t=-\frac{6}{25}
To find equation solutions, solve 2t-3=0 and 25t+6=0.
50t^{2}-63t-18=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(-63\right)±\sqrt{\left(-63\right)^{2}-4\times 50\left(-18\right)}}{2\times 50}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 50 for a, -63 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-63\right)±\sqrt{3969-4\times 50\left(-18\right)}}{2\times 50}
Square -63.
t=\frac{-\left(-63\right)±\sqrt{3969-200\left(-18\right)}}{2\times 50}
Multiply -4 times 50.
t=\frac{-\left(-63\right)±\sqrt{3969+3600}}{2\times 50}
Multiply -200 times -18.
t=\frac{-\left(-63\right)±\sqrt{7569}}{2\times 50}
Add 3969 to 3600.
t=\frac{-\left(-63\right)±87}{2\times 50}
Take the square root of 7569.
t=\frac{63±87}{2\times 50}
The opposite of -63 is 63.
t=\frac{63±87}{100}
Multiply 2 times 50.
t=\frac{150}{100}
Now solve the equation t=\frac{63±87}{100} when ± is plus. Add 63 to 87.
t=\frac{3}{2}
Reduce the fraction \frac{150}{100} to lowest terms by extracting and canceling out 50.
t=-\frac{24}{100}
Now solve the equation t=\frac{63±87}{100} when ± is minus. Subtract 87 from 63.
t=-\frac{6}{25}
Reduce the fraction \frac{-24}{100} to lowest terms by extracting and canceling out 4.
t=\frac{3}{2} t=-\frac{6}{25}
The equation is now solved.
50t^{2}-63t-18=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.
50t^{2}-63t-18-\left(-18\right)=-\left(-18\right)
Add 18 to both sides of the equation.
50t^{2}-63t=-\left(-18\right)
Subtracting -18 from itself leaves 0.
50t^{2}-63t=18
Subtract -18 from 0.
\frac{50t^{2}-63t}{50}=\frac{18}{50}
Divide both sides by 50.
t^{2}-\frac{63}{50}t=\frac{18}{50}
Dividing by 50 undoes the multiplication by 50.
t^{2}-\frac{63}{50}t=\frac{9}{25}
Reduce the fraction \frac{18}{50} to lowest terms by extracting and canceling out 2.
t^{2}-\frac{63}{50}t+\left(-\frac{63}{100}\right)^{2}=\frac{9}{25}+\left(-\frac{63}{100}\right)^{2}
Divide -\frac{63}{50}, the coefficient of the x term, by 2 to get -\frac{63}{100}. Then add the square of -\frac{63}{100} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{63}{50}t+\frac{3969}{10000}=\frac{9}{25}+\frac{3969}{10000}
Square -\frac{63}{100} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{63}{50}t+\frac{3969}{10000}=\frac{7569}{10000}
Add \frac{9}{25} to \frac{3969}{10000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{63}{100}\right)^{2}=\frac{7569}{10000}
Factor t^{2}-\frac{63}{50}t+\frac{3969}{10000}. 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-\frac{63}{100}\right)^{2}}=\sqrt{\frac{7569}{10000}}
Take the square root of both sides of the equation.
t-\frac{63}{100}=\frac{87}{100} t-\frac{63}{100}=-\frac{87}{100}
Simplify.
t=\frac{3}{2} t=-\frac{6}{25}
Add \frac{63}{100} to both sides of the equation.
x ^ 2 -\frac{63}{50}x -\frac{9}{25} = 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.This is achieved by dividing both sides of the equation by 50
r + s = \frac{63}{50} rs = -\frac{9}{25}
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 = \frac{63}{100} - u s = \frac{63}{100} + u
Two numbers r and s sum up to \frac{63}{50} exactly when the average of the two numbers is \frac{1}{2}*\frac{63}{50} = \frac{63}{100}. 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>
(\frac{63}{100} - u) (\frac{63}{100} + u) = -\frac{9}{25}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{25}
\frac{3969}{10000} - u^2 = -\frac{9}{25}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{9}{25}-\frac{3969}{10000} = -\frac{7569}{10000}
Simplify the expression by subtracting \frac{3969}{10000} on both sides
u^2 = \frac{7569}{10000} u = \pm\sqrt{\frac{7569}{10000}} = \pm \frac{87}{100}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{63}{100} - \frac{87}{100} = -0.240 s = \frac{63}{100} + \frac{87}{100} = 1.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.