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2x^{2}+x-105=0
Subtract 105 from both sides.
a+b=1 ab=2\left(-105\right)=-210
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-105. 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 positive, the positive number has greater absolute value than the negative. 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=-14 b=15
The solution is the pair that gives sum 1.
\left(2x^{2}-14x\right)+\left(15x-105\right)
Rewrite 2x^{2}+x-105 as \left(2x^{2}-14x\right)+\left(15x-105\right).
2x\left(x-7\right)+15\left(x-7\right)
Factor out 2x in the first and 15 in the second group.
\left(x-7\right)\left(2x+15\right)
Factor out common term x-7 by using distributive property.
x=7 x=-\frac{15}{2}
To find equation solutions, solve x-7=0 and 2x+15=0.
2x^{2}+x=105
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.
2x^{2}+x-105=105-105
Subtract 105 from both sides of the equation.
2x^{2}+x-105=0
Subtracting 105 from itself leaves 0.
x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-105\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\times 2\left(-105\right)}}{2\times 2}
Square 1.
x=\frac{-1±\sqrt{1-8\left(-105\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-1±\sqrt{1+840}}{2\times 2}
Multiply -8 times -105.
x=\frac{-1±\sqrt{841}}{2\times 2}
Add 1 to 840.
x=\frac{-1±29}{2\times 2}
Take the square root of 841.
x=\frac{-1±29}{4}
Multiply 2 times 2.
x=\frac{28}{4}
Now solve the equation x=\frac{-1±29}{4} when ± is plus. Add -1 to 29.
x=7
Divide 28 by 4.
x=-\frac{30}{4}
Now solve the equation x=\frac{-1±29}{4} when ± is minus. Subtract 29 from -1.
x=-\frac{15}{2}
Reduce the fraction \frac{-30}{4} to lowest terms by extracting and canceling out 2.
x=7 x=-\frac{15}{2}
The equation is now solved.
2x^{2}+x=105
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{2x^{2}+x}{2}=\frac{105}{2}
Divide both sides by 2.
x^{2}+\frac{1}{2}x=\frac{105}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{105}{2}+\left(\frac{1}{4}\right)^{2}
Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{105}{2}+\frac{1}{16}
Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{841}{16}
Add \frac{105}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{4}\right)^{2}=\frac{841}{16}
Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. 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}{4}\right)^{2}}=\sqrt{\frac{841}{16}}
Take the square root of both sides of the equation.
x+\frac{1}{4}=\frac{29}{4} x+\frac{1}{4}=-\frac{29}{4}
Simplify.
x=7 x=-\frac{15}{2}
Subtract \frac{1}{4} from both sides of the equation.