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6\left(v^{2}-3v-10\right)
Factor out 6.
a+b=-3 ab=1\left(-10\right)=-10
Consider v^{2}-3v-10. Factor the expression by grouping. First, the expression needs to be rewritten as v^{2}+av+bv-10. To find a and b, set up a system to be solved.
1,-10 2,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.
1-10=-9 2-5=-3
Calculate the sum for each pair.
a=-5 b=2
The solution is the pair that gives sum -3.
\left(v^{2}-5v\right)+\left(2v-10\right)
Rewrite v^{2}-3v-10 as \left(v^{2}-5v\right)+\left(2v-10\right).
v\left(v-5\right)+2\left(v-5\right)
Factor out v in the first and 2 in the second group.
\left(v-5\right)\left(v+2\right)
Factor out common term v-5 by using distributive property.
6\left(v-5\right)\left(v+2\right)
Rewrite the complete factored expression.
6v^{2}-18v-60=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.
v=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 6\left(-60\right)}}{2\times 6}
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.
v=\frac{-\left(-18\right)±\sqrt{324-4\times 6\left(-60\right)}}{2\times 6}
Square -18.
v=\frac{-\left(-18\right)±\sqrt{324-24\left(-60\right)}}{2\times 6}
Multiply -4 times 6.
v=\frac{-\left(-18\right)±\sqrt{324+1440}}{2\times 6}
Multiply -24 times -60.
v=\frac{-\left(-18\right)±\sqrt{1764}}{2\times 6}
Add 324 to 1440.
v=\frac{-\left(-18\right)±42}{2\times 6}
Take the square root of 1764.
v=\frac{18±42}{2\times 6}
The opposite of -18 is 18.
v=\frac{18±42}{12}
Multiply 2 times 6.
v=\frac{60}{12}
Now solve the equation v=\frac{18±42}{12} when ± is plus. Add 18 to 42.
v=5
Divide 60 by 12.
v=-\frac{24}{12}
Now solve the equation v=\frac{18±42}{12} when ± is minus. Subtract 42 from 18.
v=-2
Divide -24 by 12.
6v^{2}-18v-60=6\left(v-5\right)\left(v-\left(-2\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -2 for x_{2}.
6v^{2}-18v-60=6\left(v-5\right)\left(v+2\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 -3x -10 = 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 = 3 rs = -10
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}{2} - u s = \frac{3}{2} + u
Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{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{3}{2} - u) (\frac{3}{2} + u) = -10
To solve for unknown quantity u, substitute these in the product equation rs = -10
\frac{9}{4} - u^2 = -10
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -10-\frac{9}{4} = -\frac{49}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{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{3}{2} - \frac{7}{2} = -2 s = \frac{3}{2} + \frac{7}{2} = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.