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2x^{2}-17x-34=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(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 2\left(-34\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -17 for b, and -34 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-17\right)±\sqrt{289-4\times 2\left(-34\right)}}{2\times 2}
Square -17.
x=\frac{-\left(-17\right)±\sqrt{289-8\left(-34\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-17\right)±\sqrt{289+272}}{2\times 2}
Multiply -8 times -34.
x=\frac{-\left(-17\right)±\sqrt{561}}{2\times 2}
Add 289 to 272.
x=\frac{17±\sqrt{561}}{2\times 2}
The opposite of -17 is 17.
x=\frac{17±\sqrt{561}}{4}
Multiply 2 times 2.
x=\frac{\sqrt{561}+17}{4}
Now solve the equation x=\frac{17±\sqrt{561}}{4} when ± is plus. Add 17 to \sqrt{561}.
x=\frac{17-\sqrt{561}}{4}
Now solve the equation x=\frac{17±\sqrt{561}}{4} when ± is minus. Subtract \sqrt{561} from 17.
x=\frac{\sqrt{561}+17}{4} x=\frac{17-\sqrt{561}}{4}
The equation is now solved.
2x^{2}-17x-34=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}-17x-34-\left(-34\right)=-\left(-34\right)
Add 34 to both sides of the equation.
2x^{2}-17x=-\left(-34\right)
Subtracting -34 from itself leaves 0.
2x^{2}-17x=34
Subtract -34 from 0.
\frac{2x^{2}-17x}{2}=\frac{34}{2}
Divide both sides by 2.
x^{2}-\frac{17}{2}x=\frac{34}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{17}{2}x=17
Divide 34 by 2.
x^{2}-\frac{17}{2}x+\left(-\frac{17}{4}\right)^{2}=17+\left(-\frac{17}{4}\right)^{2}
Divide -\frac{17}{2}, the coefficient of the x term, by 2 to get -\frac{17}{4}. Then add the square of -\frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{17}{2}x+\frac{289}{16}=17+\frac{289}{16}
Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{17}{2}x+\frac{289}{16}=\frac{561}{16}
Add 17 to \frac{289}{16}.
\left(x-\frac{17}{4}\right)^{2}=\frac{561}{16}
Factor x^{2}-\frac{17}{2}x+\frac{289}{16}. 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{17}{4}\right)^{2}}=\sqrt{\frac{561}{16}}
Take the square root of both sides of the equation.
x-\frac{17}{4}=\frac{\sqrt{561}}{4} x-\frac{17}{4}=-\frac{\sqrt{561}}{4}
Simplify.
x=\frac{\sqrt{561}+17}{4} x=\frac{17-\sqrt{561}}{4}
Add \frac{17}{4} to both sides of the equation.
x ^ 2 -\frac{17}{2}x -17 = 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 = \frac{17}{2} rs = -17
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{17}{4} - u s = \frac{17}{4} + u
Two numbers r and s sum up to \frac{17}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{2} = \frac{17}{4}. 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{17}{4} - u) (\frac{17}{4} + u) = -17
To solve for unknown quantity u, substitute these in the product equation rs = -17
\frac{289}{16} - u^2 = -17
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -17-\frac{289}{16} = -\frac{561}{16}
Simplify the expression by subtracting \frac{289}{16} on both sides
u^2 = \frac{561}{16} u = \pm\sqrt{\frac{561}{16}} = \pm \frac{\sqrt{561}}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{17}{4} - \frac{\sqrt{561}}{4} = -1.671 s = \frac{17}{4} + \frac{\sqrt{561}}{4} = 10.171
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.