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3x^{2}+6x-10=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{-6±\sqrt{6^{2}-4\times 3\left(-10\right)}}{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{-6±\sqrt{36-4\times 3\left(-10\right)}}{2\times 3}
Square 6.
x=\frac{-6±\sqrt{36-12\left(-10\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-6±\sqrt{36+120}}{2\times 3}
Multiply -12 times -10.
x=\frac{-6±\sqrt{156}}{2\times 3}
Add 36 to 120.
x=\frac{-6±2\sqrt{39}}{2\times 3}
Take the square root of 156.
x=\frac{-6±2\sqrt{39}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{39}-6}{6}
Now solve the equation x=\frac{-6±2\sqrt{39}}{6} when ± is plus. Add -6 to 2\sqrt{39}.
x=\frac{\sqrt{39}}{3}-1
Divide -6+2\sqrt{39} by 6.
x=\frac{-2\sqrt{39}-6}{6}
Now solve the equation x=\frac{-6±2\sqrt{39}}{6} when ± is minus. Subtract 2\sqrt{39} from -6.
x=-\frac{\sqrt{39}}{3}-1
Divide -6-2\sqrt{39} by 6.
3x^{2}+6x-10=3\left(x-\left(\frac{\sqrt{39}}{3}-1\right)\right)\left(x-\left(-\frac{\sqrt{39}}{3}-1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1+\frac{\sqrt{39}}{3} for x_{1} and -1-\frac{\sqrt{39}}{3} for x_{2}.
x ^ 2 +2x -\frac{10}{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 = -2 rs = -\frac{10}{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 = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. 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>
(-1 - u) (-1 + u) = -\frac{10}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{10}{3}
1 - u^2 = -\frac{10}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{10}{3}-1 = -\frac{13}{3}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{13}{3} u = \pm\sqrt{\frac{13}{3}} = \pm \frac{\sqrt{13}}{\sqrt{3}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \frac{\sqrt{13}}{\sqrt{3}} = -3.082 s = -1 + \frac{\sqrt{13}}{\sqrt{3}} = 1.082
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.