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a+b=2 ab=-840
To solve the equation, factor m^{2}+2m-840 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.
-1,840 -2,420 -3,280 -4,210 -5,168 -6,140 -7,120 -8,105 -10,84 -12,70 -14,60 -15,56 -20,42 -21,40 -24,35 -28,30
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -840.
-1+840=839 -2+420=418 -3+280=277 -4+210=206 -5+168=163 -6+140=134 -7+120=113 -8+105=97 -10+84=74 -12+70=58 -14+60=46 -15+56=41 -20+42=22 -21+40=19 -24+35=11 -28+30=2
Calculate the sum for each pair.
a=-28 b=30
The solution is the pair that gives sum 2.
\left(m-28\right)\left(m+30\right)
Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.
m=28 m=-30
To find equation solutions, solve m-28=0 and m+30=0.
a+b=2 ab=1\left(-840\right)=-840
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm-840. To find a and b, set up a system to be solved.
-1,840 -2,420 -3,280 -4,210 -5,168 -6,140 -7,120 -8,105 -10,84 -12,70 -14,60 -15,56 -20,42 -21,40 -24,35 -28,30
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -840.
-1+840=839 -2+420=418 -3+280=277 -4+210=206 -5+168=163 -6+140=134 -7+120=113 -8+105=97 -10+84=74 -12+70=58 -14+60=46 -15+56=41 -20+42=22 -21+40=19 -24+35=11 -28+30=2
Calculate the sum for each pair.
a=-28 b=30
The solution is the pair that gives sum 2.
\left(m^{2}-28m\right)+\left(30m-840\right)
Rewrite m^{2}+2m-840 as \left(m^{2}-28m\right)+\left(30m-840\right).
m\left(m-28\right)+30\left(m-28\right)
Factor out m in the first and 30 in the second group.
\left(m-28\right)\left(m+30\right)
Factor out common term m-28 by using distributive property.
m=28 m=-30
To find equation solutions, solve m-28=0 and m+30=0.
m^{2}+2m-840=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.
m=\frac{-2±\sqrt{2^{2}-4\left(-840\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -840 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
m=\frac{-2±\sqrt{4-4\left(-840\right)}}{2}
Square 2.
m=\frac{-2±\sqrt{4+3360}}{2}
Multiply -4 times -840.
m=\frac{-2±\sqrt{3364}}{2}
Add 4 to 3360.
m=\frac{-2±58}{2}
Take the square root of 3364.
m=\frac{56}{2}
Now solve the equation m=\frac{-2±58}{2} when ± is plus. Add -2 to 58.
m=28
Divide 56 by 2.
m=-\frac{60}{2}
Now solve the equation m=\frac{-2±58}{2} when ± is minus. Subtract 58 from -2.
m=-30
Divide -60 by 2.
m=28 m=-30
The equation is now solved.
m^{2}+2m-840=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.
m^{2}+2m-840-\left(-840\right)=-\left(-840\right)
Add 840 to both sides of the equation.
m^{2}+2m=-\left(-840\right)
Subtracting -840 from itself leaves 0.
m^{2}+2m=840
Subtract -840 from 0.
m^{2}+2m+1^{2}=840+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.
m^{2}+2m+1=840+1
Square 1.
m^{2}+2m+1=841
Add 840 to 1.
\left(m+1\right)^{2}=841
Factor m^{2}+2m+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(m+1\right)^{2}}=\sqrt{841}
Take the square root of both sides of the equation.
m+1=29 m+1=-29
Simplify.
m=28 m=-30
Subtract 1 from both sides of the equation.
x ^ 2 +2x -840 = 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 = -2 rs = -840
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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-1 - u) (-1 + u) = -840
To solve for unknown quantity u, substitute these in the product equation rs = -840
1 - u^2 = -840
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -840-1 = -841
Simplify the expression by subtracting 1 on both sides
u^2 = 841 u = \pm\sqrt{841} = \pm 29
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - 29 = -30 s = -1 + 29 = 28
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.