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6x^{2}+12x-5=0
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{-12±\sqrt{12^{2}-4\times 6\left(-5\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 12 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-12±\sqrt{144-4\times 6\left(-5\right)}}{2\times 6}
Square 12.
x=\frac{-12±\sqrt{144-24\left(-5\right)}}{2\times 6}
Multiply -4 times 6.
x=\frac{-12±\sqrt{144+120}}{2\times 6}
Multiply -24 times -5.
x=\frac{-12±\sqrt{264}}{2\times 6}
Add 144 to 120.
x=\frac{-12±2\sqrt{66}}{2\times 6}
Take the square root of 264.
x=\frac{-12±2\sqrt{66}}{12}
Multiply 2 times 6.
x=\frac{2\sqrt{66}-12}{12}
Now solve the equation x=\frac{-12±2\sqrt{66}}{12} when ± is plus. Add -12 to 2\sqrt{66}.
x=\frac{\sqrt{66}}{6}-1
Divide -12+2\sqrt{66} by 12.
x=\frac{-2\sqrt{66}-12}{12}
Now solve the equation x=\frac{-12±2\sqrt{66}}{12} when ± is minus. Subtract 2\sqrt{66} from -12.
x=-\frac{\sqrt{66}}{6}-1
Divide -12-2\sqrt{66} by 12.
x=\frac{\sqrt{66}}{6}-1 x=-\frac{\sqrt{66}}{6}-1
The equation is now solved.
6x^{2}+12x-5=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
6x^{2}+12x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
6x^{2}+12x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
6x^{2}+12x=5
Subtract -5 from 0.
\frac{6x^{2}+12x}{6}=\frac{5}{6}
Divide both sides by 6.
x^{2}+\frac{12}{6}x=\frac{5}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}+2x=\frac{5}{6}
Divide 12 by 6.
x^{2}+2x+1^{2}=\frac{5}{6}+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=\frac{5}{6}+1
Square 1.
x^{2}+2x+1=\frac{11}{6}
Add \frac{5}{6} to 1.
\left(x+1\right)^{2}=\frac{11}{6}
Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{11}{6}}
Take the square root of both sides of the equation.
x+1=\frac{\sqrt{66}}{6} x+1=-\frac{\sqrt{66}}{6}
Simplify.
x=\frac{\sqrt{66}}{6}-1 x=-\frac{\sqrt{66}}{6}-1
Subtract 1 from both sides of the equation.
x ^ 2 +2x -\frac{5}{6} = 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 6
r + s = -2 rs = -\frac{5}{6}
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{5}{6}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{6}
1 - u^2 = -\frac{5}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{6}-1 = -\frac{11}{6}
Simplify the expression by subtracting 1 on both sides
u^2 = \frac{11}{6} u = \pm\sqrt{\frac{11}{6}} = \pm \frac{\sqrt{11}}{\sqrt{6}}
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{11}}{\sqrt{6}} = -2.354 s = -1 + \frac{\sqrt{11}}{\sqrt{6}} = 0.354
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.