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3x^{2}-15x+9=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(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\times 9}}{2\times 3}
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(-15\right)±\sqrt{225-4\times 3\times 9}}{2\times 3}
Square -15.
x=\frac{-\left(-15\right)±\sqrt{225-12\times 9}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-15\right)±\sqrt{225-108}}{2\times 3}
Multiply -12 times 9.
x=\frac{-\left(-15\right)±\sqrt{117}}{2\times 3}
Add 225 to -108.
x=\frac{-\left(-15\right)±3\sqrt{13}}{2\times 3}
Take the square root of 117.
x=\frac{15±3\sqrt{13}}{2\times 3}
The opposite of -15 is 15.
x=\frac{15±3\sqrt{13}}{6}
Multiply 2 times 3.
x=\frac{3\sqrt{13}+15}{6}
Now solve the equation x=\frac{15±3\sqrt{13}}{6} when ± is plus. Add 15 to 3\sqrt{13}.
x=\frac{\sqrt{13}+5}{2}
Divide 15+3\sqrt{13} by 6.
x=\frac{15-3\sqrt{13}}{6}
Now solve the equation x=\frac{15±3\sqrt{13}}{6} when ± is minus. Subtract 3\sqrt{13} from 15.
x=\frac{5-\sqrt{13}}{2}
Divide 15-3\sqrt{13} by 6.
3x^{2}-15x+9=3\left(x-\frac{\sqrt{13}+5}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5+\sqrt{13}}{2} for x_{1} and \frac{5-\sqrt{13}}{2} for x_{2}.
x ^ 2 -5x +3 = 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 3
r + s = 5 rs = 3
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{5}{2} - u s = \frac{5}{2} + u
Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{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{5}{2} - u) (\frac{5}{2} + u) = 3
To solve for unknown quantity u, substitute these in the product equation rs = 3
\frac{25}{4} - u^2 = 3
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 3-\frac{25}{4} = -\frac{13}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{13}{4} u = \pm\sqrt{\frac{13}{4}} = \pm \frac{\sqrt{13}}{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{5}{2} - \frac{\sqrt{13}}{2} = 0.697 s = \frac{5}{2} + \frac{\sqrt{13}}{2} = 4.303
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.