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5\left(-x^{2}-5x+14\right)
Factor out 5.
a+b=-5 ab=-14=-14
Consider -x^{2}-5x+14. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+14. To find a and b, set up a system to be solved.
1,-14 2,-7
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 -14.
1-14=-13 2-7=-5
Calculate the sum for each pair.
a=2 b=-7
The solution is the pair that gives sum -5.
\left(-x^{2}+2x\right)+\left(-7x+14\right)
Rewrite -x^{2}-5x+14 as \left(-x^{2}+2x\right)+\left(-7x+14\right).
x\left(-x+2\right)+7\left(-x+2\right)
Factor out x in the first and 7 in the second group.
\left(-x+2\right)\left(x+7\right)
Factor out common term -x+2 by using distributive property.
5\left(-x+2\right)\left(x+7\right)
Rewrite the complete factored expression.
-5x^{2}-25x+70=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{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\left(-5\right)\times 70}}{2\left(-5\right)}
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{-\left(-25\right)±\sqrt{625-4\left(-5\right)\times 70}}{2\left(-5\right)}
Square -25.
x=\frac{-\left(-25\right)±\sqrt{625+20\times 70}}{2\left(-5\right)}
Multiply -4 times -5.
x=\frac{-\left(-25\right)±\sqrt{625+1400}}{2\left(-5\right)}
Multiply 20 times 70.
x=\frac{-\left(-25\right)±\sqrt{2025}}{2\left(-5\right)}
Add 625 to 1400.
x=\frac{-\left(-25\right)±45}{2\left(-5\right)}
Take the square root of 2025.
x=\frac{25±45}{2\left(-5\right)}
The opposite of -25 is 25.
x=\frac{25±45}{-10}
Multiply 2 times -5.
x=\frac{70}{-10}
Now solve the equation x=\frac{25±45}{-10} when ± is plus. Add 25 to 45.
x=-7
Divide 70 by -10.
x=-\frac{20}{-10}
Now solve the equation x=\frac{25±45}{-10} when ± is minus. Subtract 45 from 25.
x=2
Divide -20 by -10.
-5x^{2}-25x+70=-5\left(x-\left(-7\right)\right)\left(x-2\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -7 for x_{1} and 2 for x_{2}.
-5x^{2}-25x+70=-5\left(x+7\right)\left(x-2\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +5x -14 = 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.
r + s = -5 rs = -14
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{5}{2} - u s = -\frac{5}{2} + u
Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{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{5}{2} - u) (-\frac{5}{2} + u) = -14
To solve for unknown quantity u, substitute these in the product equation rs = -14
\frac{25}{4} - u^2 = -14
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -14-\frac{25}{4} = -\frac{81}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{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{5}{2} - \frac{9}{2} = -7 s = -\frac{5}{2} + \frac{9}{2} = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.