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a+b=27 ab=7\left(-40\right)=-280
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-40. To find a and b, set up a system to be solved.
-1,280 -2,140 -4,70 -5,56 -7,40 -8,35 -10,28 -14,20
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 -280.
-1+280=279 -2+140=138 -4+70=66 -5+56=51 -7+40=33 -8+35=27 -10+28=18 -14+20=6
Calculate the sum for each pair.
a=-8 b=35
The solution is the pair that gives sum 27.
\left(7x^{2}-8x\right)+\left(35x-40\right)
Rewrite 7x^{2}+27x-40 as \left(7x^{2}-8x\right)+\left(35x-40\right).
x\left(7x-8\right)+5\left(7x-8\right)
Factor out x in the first and 5 in the second group.
\left(7x-8\right)\left(x+5\right)
Factor out common term 7x-8 by using distributive property.
x=\frac{8}{7} x=-5
To find equation solutions, solve 7x-8=0 and x+5=0.
7x^{2}+27x-40=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{-27±\sqrt{27^{2}-4\times 7\left(-40\right)}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 27 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-27±\sqrt{729-4\times 7\left(-40\right)}}{2\times 7}
Square 27.
x=\frac{-27±\sqrt{729-28\left(-40\right)}}{2\times 7}
Multiply -4 times 7.
x=\frac{-27±\sqrt{729+1120}}{2\times 7}
Multiply -28 times -40.
x=\frac{-27±\sqrt{1849}}{2\times 7}
Add 729 to 1120.
x=\frac{-27±43}{2\times 7}
Take the square root of 1849.
x=\frac{-27±43}{14}
Multiply 2 times 7.
x=\frac{16}{14}
Now solve the equation x=\frac{-27±43}{14} when ± is plus. Add -27 to 43.
x=\frac{8}{7}
Reduce the fraction \frac{16}{14} to lowest terms by extracting and canceling out 2.
x=-\frac{70}{14}
Now solve the equation x=\frac{-27±43}{14} when ± is minus. Subtract 43 from -27.
x=-5
Divide -70 by 14.
x=\frac{8}{7} x=-5
The equation is now solved.
7x^{2}+27x-40=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.
7x^{2}+27x-40-\left(-40\right)=-\left(-40\right)
Add 40 to both sides of the equation.
7x^{2}+27x=-\left(-40\right)
Subtracting -40 from itself leaves 0.
7x^{2}+27x=40
Subtract -40 from 0.
\frac{7x^{2}+27x}{7}=\frac{40}{7}
Divide both sides by 7.
x^{2}+\frac{27}{7}x=\frac{40}{7}
Dividing by 7 undoes the multiplication by 7.
x^{2}+\frac{27}{7}x+\left(\frac{27}{14}\right)^{2}=\frac{40}{7}+\left(\frac{27}{14}\right)^{2}
Divide \frac{27}{7}, the coefficient of the x term, by 2 to get \frac{27}{14}. Then add the square of \frac{27}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{27}{7}x+\frac{729}{196}=\frac{40}{7}+\frac{729}{196}
Square \frac{27}{14} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{27}{7}x+\frac{729}{196}=\frac{1849}{196}
Add \frac{40}{7} to \frac{729}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{27}{14}\right)^{2}=\frac{1849}{196}
Factor x^{2}+\frac{27}{7}x+\frac{729}{196}. 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+\frac{27}{14}\right)^{2}}=\sqrt{\frac{1849}{196}}
Take the square root of both sides of the equation.
x+\frac{27}{14}=\frac{43}{14} x+\frac{27}{14}=-\frac{43}{14}
Simplify.
x=\frac{8}{7} x=-5
Subtract \frac{27}{14} from both sides of the equation.