Skip to main content
Factor
Tick mark Image
Evaluate
Tick mark Image
Graph

Similar Problems from Web Search

Share

5x^{2}+6x-7=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 5\left(-7\right)}}{2\times 5}
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 5\left(-7\right)}}{2\times 5}
Square 6.
x=\frac{-6±\sqrt{36-20\left(-7\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-6±\sqrt{36+140}}{2\times 5}
Multiply -20 times -7.
x=\frac{-6±\sqrt{176}}{2\times 5}
Add 36 to 140.
x=\frac{-6±4\sqrt{11}}{2\times 5}
Take the square root of 176.
x=\frac{-6±4\sqrt{11}}{10}
Multiply 2 times 5.
x=\frac{4\sqrt{11}-6}{10}
Now solve the equation x=\frac{-6±4\sqrt{11}}{10} when ± is plus. Add -6 to 4\sqrt{11}.
x=\frac{2\sqrt{11}-3}{5}
Divide -6+4\sqrt{11} by 10.
x=\frac{-4\sqrt{11}-6}{10}
Now solve the equation x=\frac{-6±4\sqrt{11}}{10} when ± is minus. Subtract 4\sqrt{11} from -6.
x=\frac{-2\sqrt{11}-3}{5}
Divide -6-4\sqrt{11} by 10.
5x^{2}+6x-7=5\left(x-\frac{2\sqrt{11}-3}{5}\right)\left(x-\frac{-2\sqrt{11}-3}{5}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+2\sqrt{11}}{5} for x_{1} and \frac{-3-2\sqrt{11}}{5} for x_{2}.
x ^ 2 +\frac{6}{5}x -\frac{7}{5} = 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 5
r + s = -\frac{6}{5} rs = -\frac{7}{5}
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{3}{5} - u s = -\frac{3}{5} + u
Two numbers r and s sum up to -\frac{6}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{6}{5} = -\frac{3}{5}. 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{3}{5} - u) (-\frac{3}{5} + u) = -\frac{7}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{5}
\frac{9}{25} - u^2 = -\frac{7}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{7}{5}-\frac{9}{25} = -\frac{44}{25}
Simplify the expression by subtracting \frac{9}{25} on both sides
u^2 = \frac{44}{25} u = \pm\sqrt{\frac{44}{25}} = \pm \frac{\sqrt{44}}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{5} - \frac{\sqrt{44}}{5} = -1.927 s = -\frac{3}{5} + \frac{\sqrt{44}}{5} = 0.727
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.