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a+b=3 ab=35\left(-2\right)=-70
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 35x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
-1,70 -2,35 -5,14 -7,10
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 -70.
-1+70=69 -2+35=33 -5+14=9 -7+10=3
Calculate the sum for each pair.
a=-7 b=10
The solution is the pair that gives sum 3.
\left(35x^{2}-7x\right)+\left(10x-2\right)
Rewrite 35x^{2}+3x-2 as \left(35x^{2}-7x\right)+\left(10x-2\right).
7x\left(5x-1\right)+2\left(5x-1\right)
Factor out 7x in the first and 2 in the second group.
\left(5x-1\right)\left(7x+2\right)
Factor out common term 5x-1 by using distributive property.
x=\frac{1}{5} x=-\frac{2}{7}
To find equation solutions, solve 5x-1=0 and 7x+2=0.
35x^{2}+3x-2=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{-3±\sqrt{3^{2}-4\times 35\left(-2\right)}}{2\times 35}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 35 for a, 3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times 35\left(-2\right)}}{2\times 35}
Square 3.
x=\frac{-3±\sqrt{9-140\left(-2\right)}}{2\times 35}
Multiply -4 times 35.
x=\frac{-3±\sqrt{9+280}}{2\times 35}
Multiply -140 times -2.
x=\frac{-3±\sqrt{289}}{2\times 35}
Add 9 to 280.
x=\frac{-3±17}{2\times 35}
Take the square root of 289.
x=\frac{-3±17}{70}
Multiply 2 times 35.
x=\frac{14}{70}
Now solve the equation x=\frac{-3±17}{70} when ± is plus. Add -3 to 17.
x=\frac{1}{5}
Reduce the fraction \frac{14}{70} to lowest terms by extracting and canceling out 14.
x=-\frac{20}{70}
Now solve the equation x=\frac{-3±17}{70} when ± is minus. Subtract 17 from -3.
x=-\frac{2}{7}
Reduce the fraction \frac{-20}{70} to lowest terms by extracting and canceling out 10.
x=\frac{1}{5} x=-\frac{2}{7}
The equation is now solved.
35x^{2}+3x-2=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.
35x^{2}+3x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
35x^{2}+3x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
35x^{2}+3x=2
Subtract -2 from 0.
\frac{35x^{2}+3x}{35}=\frac{2}{35}
Divide both sides by 35.
x^{2}+\frac{3}{35}x=\frac{2}{35}
Dividing by 35 undoes the multiplication by 35.
x^{2}+\frac{3}{35}x+\left(\frac{3}{70}\right)^{2}=\frac{2}{35}+\left(\frac{3}{70}\right)^{2}
Divide \frac{3}{35}, the coefficient of the x term, by 2 to get \frac{3}{70}. Then add the square of \frac{3}{70} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{3}{35}x+\frac{9}{4900}=\frac{2}{35}+\frac{9}{4900}
Square \frac{3}{70} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{3}{35}x+\frac{9}{4900}=\frac{289}{4900}
Add \frac{2}{35} to \frac{9}{4900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{70}\right)^{2}=\frac{289}{4900}
Factor x^{2}+\frac{3}{35}x+\frac{9}{4900}. 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{3}{70}\right)^{2}}=\sqrt{\frac{289}{4900}}
Take the square root of both sides of the equation.
x+\frac{3}{70}=\frac{17}{70} x+\frac{3}{70}=-\frac{17}{70}
Simplify.
x=\frac{1}{5} x=-\frac{2}{7}
Subtract \frac{3}{70} from both sides of the equation.