Solve for x
x=\frac{\sqrt{3769}-47}{40}\approx 0.359804548
x=\frac{-\sqrt{3769}-47}{40}\approx -2.709804548
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40x^{2}+94x-39=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{-94±\sqrt{94^{2}-4\times 40\left(-39\right)}}{2\times 40}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, 94 for b, and -39 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-94±\sqrt{8836-4\times 40\left(-39\right)}}{2\times 40}
Square 94.
x=\frac{-94±\sqrt{8836-160\left(-39\right)}}{2\times 40}
Multiply -4 times 40.
x=\frac{-94±\sqrt{8836+6240}}{2\times 40}
Multiply -160 times -39.
x=\frac{-94±\sqrt{15076}}{2\times 40}
Add 8836 to 6240.
x=\frac{-94±2\sqrt{3769}}{2\times 40}
Take the square root of 15076.
x=\frac{-94±2\sqrt{3769}}{80}
Multiply 2 times 40.
x=\frac{2\sqrt{3769}-94}{80}
Now solve the equation x=\frac{-94±2\sqrt{3769}}{80} when ± is plus. Add -94 to 2\sqrt{3769}.
x=\frac{\sqrt{3769}-47}{40}
Divide -94+2\sqrt{3769} by 80.
x=\frac{-2\sqrt{3769}-94}{80}
Now solve the equation x=\frac{-94±2\sqrt{3769}}{80} when ± is minus. Subtract 2\sqrt{3769} from -94.
x=\frac{-\sqrt{3769}-47}{40}
Divide -94-2\sqrt{3769} by 80.
x=\frac{\sqrt{3769}-47}{40} x=\frac{-\sqrt{3769}-47}{40}
The equation is now solved.
40x^{2}+94x-39=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.
40x^{2}+94x-39-\left(-39\right)=-\left(-39\right)
Add 39 to both sides of the equation.
40x^{2}+94x=-\left(-39\right)
Subtracting -39 from itself leaves 0.
40x^{2}+94x=39
Subtract -39 from 0.
\frac{40x^{2}+94x}{40}=\frac{39}{40}
Divide both sides by 40.
x^{2}+\frac{94}{40}x=\frac{39}{40}
Dividing by 40 undoes the multiplication by 40.
x^{2}+\frac{47}{20}x=\frac{39}{40}
Reduce the fraction \frac{94}{40} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{47}{20}x+\left(\frac{47}{40}\right)^{2}=\frac{39}{40}+\left(\frac{47}{40}\right)^{2}
Divide \frac{47}{20}, the coefficient of the x term, by 2 to get \frac{47}{40}. Then add the square of \frac{47}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{47}{20}x+\frac{2209}{1600}=\frac{39}{40}+\frac{2209}{1600}
Square \frac{47}{40} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{47}{20}x+\frac{2209}{1600}=\frac{3769}{1600}
Add \frac{39}{40} to \frac{2209}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{47}{40}\right)^{2}=\frac{3769}{1600}
Factor x^{2}+\frac{47}{20}x+\frac{2209}{1600}. 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{47}{40}\right)^{2}}=\sqrt{\frac{3769}{1600}}
Take the square root of both sides of the equation.
x+\frac{47}{40}=\frac{\sqrt{3769}}{40} x+\frac{47}{40}=-\frac{\sqrt{3769}}{40}
Simplify.
x=\frac{\sqrt{3769}-47}{40} x=\frac{-\sqrt{3769}-47}{40}
Subtract \frac{47}{40} from both sides of the equation.
x ^ 2 +\frac{47}{20}x -\frac{39}{40} = 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 40
r + s = -\frac{47}{20} rs = -\frac{39}{40}
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{47}{40} - u s = -\frac{47}{40} + u
Two numbers r and s sum up to -\frac{47}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{47}{20} = -\frac{47}{40}. 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{47}{40} - u) (-\frac{47}{40} + u) = -\frac{39}{40}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{39}{40}
\frac{2209}{1600} - u^2 = -\frac{39}{40}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{39}{40}-\frac{2209}{1600} = -\frac{3769}{1600}
Simplify the expression by subtracting \frac{2209}{1600} on both sides
u^2 = \frac{3769}{1600} u = \pm\sqrt{\frac{3769}{1600}} = \pm \frac{\sqrt{3769}}{40}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{47}{40} - \frac{\sqrt{3769}}{40} = -2.710 s = -\frac{47}{40} + \frac{\sqrt{3769}}{40} = 0.360
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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