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x^{2}+4x+3-5040=0
Subtract 5040 from both sides.
x^{2}+4x-5037=0
Subtract 5040 from 3 to get -5037.
a+b=4 ab=-5037
To solve the equation, factor x^{2}+4x-5037 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
-1,5037 -3,1679 -23,219 -69,73
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 -5037.
-1+5037=5036 -3+1679=1676 -23+219=196 -69+73=4
Calculate the sum for each pair.
a=-69 b=73
The solution is the pair that gives sum 4.
\left(x-69\right)\left(x+73\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=69 x=-73
To find equation solutions, solve x-69=0 and x+73=0.
x^{2}+4x+3-5040=0
Subtract 5040 from both sides.
x^{2}+4x-5037=0
Subtract 5040 from 3 to get -5037.
a+b=4 ab=1\left(-5037\right)=-5037
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-5037. To find a and b, set up a system to be solved.
-1,5037 -3,1679 -23,219 -69,73
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 -5037.
-1+5037=5036 -3+1679=1676 -23+219=196 -69+73=4
Calculate the sum for each pair.
a=-69 b=73
The solution is the pair that gives sum 4.
\left(x^{2}-69x\right)+\left(73x-5037\right)
Rewrite x^{2}+4x-5037 as \left(x^{2}-69x\right)+\left(73x-5037\right).
x\left(x-69\right)+73\left(x-69\right)
Factor out x in the first and 73 in the second group.
\left(x-69\right)\left(x+73\right)
Factor out common term x-69 by using distributive property.
x=69 x=-73
To find equation solutions, solve x-69=0 and x+73=0.
x^{2}+4x+3=5040
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^{2}+4x+3-5040=5040-5040
Subtract 5040 from both sides of the equation.
x^{2}+4x+3-5040=0
Subtracting 5040 from itself leaves 0.
x^{2}+4x-5037=0
Subtract 5040 from 3.
x=\frac{-4±\sqrt{4^{2}-4\left(-5037\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -5037 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\left(-5037\right)}}{2}
Square 4.
x=\frac{-4±\sqrt{16+20148}}{2}
Multiply -4 times -5037.
x=\frac{-4±\sqrt{20164}}{2}
Add 16 to 20148.
x=\frac{-4±142}{2}
Take the square root of 20164.
x=\frac{138}{2}
Now solve the equation x=\frac{-4±142}{2} when ± is plus. Add -4 to 142.
x=69
Divide 138 by 2.
x=-\frac{146}{2}
Now solve the equation x=\frac{-4±142}{2} when ± is minus. Subtract 142 from -4.
x=-73
Divide -146 by 2.
x=69 x=-73
The equation is now solved.
x^{2}+4x+3=5040
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.
x^{2}+4x+3-3=5040-3
Subtract 3 from both sides of the equation.
x^{2}+4x=5040-3
Subtracting 3 from itself leaves 0.
x^{2}+4x=5037
Subtract 3 from 5040.
x^{2}+4x+2^{2}=5037+2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+4x+4=5037+4
Square 2.
x^{2}+4x+4=5041
Add 5037 to 4.
\left(x+2\right)^{2}=5041
Factor x^{2}+4x+4. 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+2\right)^{2}}=\sqrt{5041}
Take the square root of both sides of the equation.
x+2=71 x+2=-71
Simplify.
x=69 x=-73
Subtract 2 from both sides of the equation.