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3n^{2}+8n-1050=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.
n=\frac{-8±\sqrt{8^{2}-4\times 3\left(-1050\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.
n=\frac{-8±\sqrt{64-4\times 3\left(-1050\right)}}{2\times 3}
Square 8.
n=\frac{-8±\sqrt{64-12\left(-1050\right)}}{2\times 3}
Multiply -4 times 3.
n=\frac{-8±\sqrt{64+12600}}{2\times 3}
Multiply -12 times -1050.
n=\frac{-8±\sqrt{12664}}{2\times 3}
Add 64 to 12600.
n=\frac{-8±2\sqrt{3166}}{2\times 3}
Take the square root of 12664.
n=\frac{-8±2\sqrt{3166}}{6}
Multiply 2 times 3.
n=\frac{2\sqrt{3166}-8}{6}
Now solve the equation n=\frac{-8±2\sqrt{3166}}{6} when ± is plus. Add -8 to 2\sqrt{3166}.
n=\frac{\sqrt{3166}-4}{3}
Divide -8+2\sqrt{3166} by 6.
n=\frac{-2\sqrt{3166}-8}{6}
Now solve the equation n=\frac{-8±2\sqrt{3166}}{6} when ± is minus. Subtract 2\sqrt{3166} from -8.
n=\frac{-\sqrt{3166}-4}{3}
Divide -8-2\sqrt{3166} by 6.
3n^{2}+8n-1050=3\left(n-\frac{\sqrt{3166}-4}{3}\right)\left(n-\frac{-\sqrt{3166}-4}{3}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-4+\sqrt{3166}}{3} for x_{1} and \frac{-4-\sqrt{3166}}{3} for x_{2}.
x ^ 2 +\frac{8}{3}x -350 = 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 = -\frac{8}{3} rs = -350
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{4}{3} - u s = -\frac{4}{3} + u
Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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{4}{3} - u) (-\frac{4}{3} + u) = -350
To solve for unknown quantity u, substitute these in the product equation rs = -350
\frac{16}{9} - u^2 = -350
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -350-\frac{16}{9} = -\frac{3166}{9}
Simplify the expression by subtracting \frac{16}{9} on both sides
u^2 = \frac{3166}{9} u = \pm\sqrt{\frac{3166}{9}} = \pm \frac{\sqrt{3166}}{3}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{4}{3} - \frac{\sqrt{3166}}{3} = -20.089 s = -\frac{4}{3} + \frac{\sqrt{3166}}{3} = 17.422
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.