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4x^{2}-17x+3=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 4\times 3}}{2\times 4}
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{289-4\times 4\times 3}}{2\times 4}
Square -17.
x=\frac{-\left(-17\right)±\sqrt{289-16\times 3}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-17\right)±\sqrt{289-48}}{2\times 4}
Multiply -16 times 3.
x=\frac{-\left(-17\right)±\sqrt{241}}{2\times 4}
Add 289 to -48.
x=\frac{17±\sqrt{241}}{2\times 4}
The opposite of -17 is 17.
x=\frac{17±\sqrt{241}}{8}
Multiply 2 times 4.
x=\frac{\sqrt{241}+17}{8}
Now solve the equation x=\frac{17±\sqrt{241}}{8} when ± is plus. Add 17 to \sqrt{241}.
x=\frac{17-\sqrt{241}}{8}
Now solve the equation x=\frac{17±\sqrt{241}}{8} when ± is minus. Subtract \sqrt{241} from 17.
4x^{2}-17x+3=4\left(x-\frac{\sqrt{241}+17}{8}\right)\left(x-\frac{17-\sqrt{241}}{8}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{17+\sqrt{241}}{8} for x_{1} and \frac{17-\sqrt{241}}{8} for x_{2}.
x ^ 2 -\frac{17}{4}x +\frac{3}{4} = 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 4
r + s = \frac{17}{4} rs = \frac{3}{4}
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}{8} - u s = \frac{17}{8} + u
Two numbers r and s sum up to \frac{17}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{17}{4} = \frac{17}{8}. 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}{8} - u) (\frac{17}{8} + u) = \frac{3}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{4}
\frac{289}{64} - u^2 = \frac{3}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{4}-\frac{289}{64} = -\frac{241}{64}
Simplify the expression by subtracting \frac{289}{64} on both sides
u^2 = \frac{241}{64} u = \pm\sqrt{\frac{241}{64}} = \pm \frac{\sqrt{241}}{8}
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}{8} - \frac{\sqrt{241}}{8} = 0.184 s = \frac{17}{8} + \frac{\sqrt{241}}{8} = 4.066
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.