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x^{2}\times 3-x-70=0
Subtract 70 from both sides.
a+b=-1 ab=3\left(-70\right)=-210
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-70. To find a and b, set up a system to be solved.
1,-210 2,-105 3,-70 5,-42 6,-35 7,-30 10,-21 14,-15
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 -210.
1-210=-209 2-105=-103 3-70=-67 5-42=-37 6-35=-29 7-30=-23 10-21=-11 14-15=-1
Calculate the sum for each pair.
a=-15 b=14
The solution is the pair that gives sum -1.
\left(3x^{2}-15x\right)+\left(14x-70\right)
Rewrite 3x^{2}-x-70 as \left(3x^{2}-15x\right)+\left(14x-70\right).
3x\left(x-5\right)+14\left(x-5\right)
Factor out 3x in the first and 14 in the second group.
\left(x-5\right)\left(3x+14\right)
Factor out common term x-5 by using distributive property.
x=5 x=-\frac{14}{3}
To find equation solutions, solve x-5=0 and 3x+14=0.
3x^{2}-x=70
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.
3x^{2}-x-70=70-70
Subtract 70 from both sides of the equation.
3x^{2}-x-70=0
Subtracting 70 from itself leaves 0.
x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-70\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-12\left(-70\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-1\right)±\sqrt{1+840}}{2\times 3}
Multiply -12 times -70.
x=\frac{-\left(-1\right)±\sqrt{841}}{2\times 3}
Add 1 to 840.
x=\frac{-\left(-1\right)±29}{2\times 3}
Take the square root of 841.
x=\frac{1±29}{2\times 3}
The opposite of -1 is 1.
x=\frac{1±29}{6}
Multiply 2 times 3.
x=\frac{30}{6}
Now solve the equation x=\frac{1±29}{6} when ± is plus. Add 1 to 29.
x=5
Divide 30 by 6.
x=-\frac{28}{6}
Now solve the equation x=\frac{1±29}{6} when ± is minus. Subtract 29 from 1.
x=-\frac{14}{3}
Reduce the fraction \frac{-28}{6} to lowest terms by extracting and canceling out 2.
x=5 x=-\frac{14}{3}
The equation is now solved.
3x^{2}-x=70
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.
\frac{3x^{2}-x}{3}=\frac{70}{3}
Divide both sides by 3.
x^{2}-\frac{1}{3}x=\frac{70}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{70}{3}+\left(-\frac{1}{6}\right)^{2}
Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{70}{3}+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{841}{36}
Add \frac{70}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{6}\right)^{2}=\frac{841}{36}
Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. 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{1}{6}\right)^{2}}=\sqrt{\frac{841}{36}}
Take the square root of both sides of the equation.
x-\frac{1}{6}=\frac{29}{6} x-\frac{1}{6}=-\frac{29}{6}
Simplify.
x=5 x=-\frac{14}{3}
Add \frac{1}{6} to both sides of the equation.