Solve for x
x = \frac{\sqrt{769} + 13}{10} \approx 4.073084925
x=\frac{13-\sqrt{769}}{10}\approx -1.473084925
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5x^{2}-13x-30=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.
x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 5\left(-30\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -13 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-13\right)±\sqrt{169-4\times 5\left(-30\right)}}{2\times 5}
Square -13.
x=\frac{-\left(-13\right)±\sqrt{169-20\left(-30\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-13\right)±\sqrt{169+600}}{2\times 5}
Multiply -20 times -30.
x=\frac{-\left(-13\right)±\sqrt{769}}{2\times 5}
Add 169 to 600.
x=\frac{13±\sqrt{769}}{2\times 5}
The opposite of -13 is 13.
x=\frac{13±\sqrt{769}}{10}
Multiply 2 times 5.
x=\frac{\sqrt{769}+13}{10}
Now solve the equation x=\frac{13±\sqrt{769}}{10} when ± is plus. Add 13 to \sqrt{769}.
x=\frac{13-\sqrt{769}}{10}
Now solve the equation x=\frac{13±\sqrt{769}}{10} when ± is minus. Subtract \sqrt{769} from 13.
x=\frac{\sqrt{769}+13}{10} x=\frac{13-\sqrt{769}}{10}
The equation is now solved.
5x^{2}-13x-30=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.
5x^{2}-13x-30-\left(-30\right)=-\left(-30\right)
Add 30 to both sides of the equation.
5x^{2}-13x=-\left(-30\right)
Subtracting -30 from itself leaves 0.
5x^{2}-13x=30
Subtract -30 from 0.
\frac{5x^{2}-13x}{5}=\frac{30}{5}
Divide both sides by 5.
x^{2}-\frac{13}{5}x=\frac{30}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{13}{5}x=6
Divide 30 by 5.
x^{2}-\frac{13}{5}x+\left(-\frac{13}{10}\right)^{2}=6+\left(-\frac{13}{10}\right)^{2}
Divide -\frac{13}{5}, the coefficient of the x term, by 2 to get -\frac{13}{10}. Then add the square of -\frac{13}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{13}{5}x+\frac{169}{100}=6+\frac{169}{100}
Square -\frac{13}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{13}{5}x+\frac{169}{100}=\frac{769}{100}
Add 6 to \frac{169}{100}.
\left(x-\frac{13}{10}\right)^{2}=\frac{769}{100}
Factor x^{2}-\frac{13}{5}x+\frac{169}{100}. 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(x-\frac{13}{10}\right)^{2}}=\sqrt{\frac{769}{100}}
Take the square root of both sides of the equation.
x-\frac{13}{10}=\frac{\sqrt{769}}{10} x-\frac{13}{10}=-\frac{\sqrt{769}}{10}
Simplify.
x=\frac{\sqrt{769}+13}{10} x=\frac{13-\sqrt{769}}{10}
Add \frac{13}{10} to both sides of the equation.
x ^ 2 -\frac{13}{5}x -6 = 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 5
r + s = \frac{13}{5} rs = -6
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{13}{10} - u s = \frac{13}{10} + u
Two numbers r and s sum up to \frac{13}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{5} = \frac{13}{10}. 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{13}{10} - u) (\frac{13}{10} + u) = -6
To solve for unknown quantity u, substitute these in the product equation rs = -6
\frac{169}{100} - u^2 = -6
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -6-\frac{169}{100} = -\frac{769}{100}
Simplify the expression by subtracting \frac{169}{100} on both sides
u^2 = \frac{769}{100} u = \pm\sqrt{\frac{769}{100}} = \pm \frac{\sqrt{769}}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{13}{10} - \frac{\sqrt{769}}{10} = -1.473 s = \frac{13}{10} + \frac{\sqrt{769}}{10} = 4.073
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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