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x^{2}+5x-0.75=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{-5±\sqrt{5^{2}-4\left(-0.75\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -0.75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\left(-0.75\right)}}{2}
Square 5.
x=\frac{-5±\sqrt{25+3}}{2}
Multiply -4 times -0.75.
x=\frac{-5±\sqrt{28}}{2}
Add 25 to 3.
x=\frac{-5±2\sqrt{7}}{2}
Take the square root of 28.
x=\frac{2\sqrt{7}-5}{2}
Now solve the equation x=\frac{-5±2\sqrt{7}}{2} when ± is plus. Add -5 to 2\sqrt{7}.
x=\sqrt{7}-\frac{5}{2}
Divide -5+2\sqrt{7} by 2.
x=\frac{-2\sqrt{7}-5}{2}
Now solve the equation x=\frac{-5±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from -5.
x=-\sqrt{7}-\frac{5}{2}
Divide -5-2\sqrt{7} by 2.
x=\sqrt{7}-\frac{5}{2} x=-\sqrt{7}-\frac{5}{2}
The equation is now solved.
x^{2}+5x-0.75=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.
x^{2}+5x-0.75-\left(-0.75\right)=-\left(-0.75\right)
Add 0.75 to both sides of the equation.
x^{2}+5x=-\left(-0.75\right)
Subtracting -0.75 from itself leaves 0.
x^{2}+5x=0.75
Subtract -0.75 from 0.
x^{2}+5x+\left(\frac{5}{2}\right)^{2}=0.75+\left(\frac{5}{2}\right)^{2}
Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+5x+\frac{25}{4}=\frac{3+25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=7
Add 0.75 to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{2}\right)^{2}=7
Factor x^{2}+5x+\frac{25}{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+\frac{5}{2}\right)^{2}}=\sqrt{7}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\sqrt{7} x+\frac{5}{2}=-\sqrt{7}
Simplify.
x=\sqrt{7}-\frac{5}{2} x=-\sqrt{7}-\frac{5}{2}
Subtract \frac{5}{2} from both sides of the equation.