Solve for x
x = \frac{\sqrt{1723} - 3}{2} \approx 19.254517581
x=\frac{-\sqrt{1723}-3}{2}\approx -22.254517581
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2x^{2}+6x-857=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{-6±\sqrt{6^{2}-4\times 2\left(-857\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -857 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times 2\left(-857\right)}}{2\times 2}
Square 6.
x=\frac{-6±\sqrt{36-8\left(-857\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-6±\sqrt{36+6856}}{2\times 2}
Multiply -8 times -857.
x=\frac{-6±\sqrt{6892}}{2\times 2}
Add 36 to 6856.
x=\frac{-6±2\sqrt{1723}}{2\times 2}
Take the square root of 6892.
x=\frac{-6±2\sqrt{1723}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{1723}-6}{4}
Now solve the equation x=\frac{-6±2\sqrt{1723}}{4} when ± is plus. Add -6 to 2\sqrt{1723}.
x=\frac{\sqrt{1723}-3}{2}
Divide -6+2\sqrt{1723} by 4.
x=\frac{-2\sqrt{1723}-6}{4}
Now solve the equation x=\frac{-6±2\sqrt{1723}}{4} when ± is minus. Subtract 2\sqrt{1723} from -6.
x=\frac{-\sqrt{1723}-3}{2}
Divide -6-2\sqrt{1723} by 4.
x=\frac{\sqrt{1723}-3}{2} x=\frac{-\sqrt{1723}-3}{2}
The equation is now solved.
2x^{2}+6x-857=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.
2x^{2}+6x-857-\left(-857\right)=-\left(-857\right)
Add 857 to both sides of the equation.
2x^{2}+6x=-\left(-857\right)
Subtracting -857 from itself leaves 0.
2x^{2}+6x=857
Subtract -857 from 0.
\frac{2x^{2}+6x}{2}=\frac{857}{2}
Divide both sides by 2.
x^{2}+\frac{6}{2}x=\frac{857}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+3x=\frac{857}{2}
Divide 6 by 2.
x^{2}+3x+\left(\frac{3}{2}\right)^{2}=\frac{857}{2}+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+3x+\frac{9}{4}=\frac{857}{2}+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+3x+\frac{9}{4}=\frac{1723}{4}
Add \frac{857}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{2}\right)^{2}=\frac{1723}{4}
Factor x^{2}+3x+\frac{9}{4}. 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{3}{2}\right)^{2}}=\sqrt{\frac{1723}{4}}
Take the square root of both sides of the equation.
x+\frac{3}{2}=\frac{\sqrt{1723}}{2} x+\frac{3}{2}=-\frac{\sqrt{1723}}{2}
Simplify.
x=\frac{\sqrt{1723}-3}{2} x=\frac{-\sqrt{1723}-3}{2}
Subtract \frac{3}{2} from both sides of the equation.
x ^ 2 +3x -\frac{857}{2} = 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 2
r + s = -3 rs = -\frac{857}{2}
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{3}{2} - u s = -\frac{3}{2} + u
Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. 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{3}{2} - u) (-\frac{3}{2} + u) = -\frac{857}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{857}{2}
\frac{9}{4} - u^2 = -\frac{857}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{857}{2}-\frac{9}{4} = -\frac{1723}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{1723}{4} u = \pm\sqrt{\frac{1723}{4}} = \pm \frac{\sqrt{1723}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{2} - \frac{\sqrt{1723}}{2} = -22.255 s = -\frac{3}{2} + \frac{\sqrt{1723}}{2} = 19.255
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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Limits
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