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7x^{2}+5x-3=54
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.
7x^{2}+5x-3-54=54-54
Subtract 54 from both sides of the equation.
7x^{2}+5x-3-54=0
Subtracting 54 from itself leaves 0.
7x^{2}+5x-57=0
Subtract 54 from -3.
x=\frac{-5±\sqrt{5^{2}-4\times 7\left(-57\right)}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 5 for b, and -57 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 7\left(-57\right)}}{2\times 7}
Square 5.
x=\frac{-5±\sqrt{25-28\left(-57\right)}}{2\times 7}
Multiply -4 times 7.
x=\frac{-5±\sqrt{25+1596}}{2\times 7}
Multiply -28 times -57.
x=\frac{-5±\sqrt{1621}}{2\times 7}
Add 25 to 1596.
x=\frac{-5±\sqrt{1621}}{14}
Multiply 2 times 7.
x=\frac{\sqrt{1621}-5}{14}
Now solve the equation x=\frac{-5±\sqrt{1621}}{14} when ± is plus. Add -5 to \sqrt{1621}.
x=\frac{-\sqrt{1621}-5}{14}
Now solve the equation x=\frac{-5±\sqrt{1621}}{14} when ± is minus. Subtract \sqrt{1621} from -5.
x=\frac{\sqrt{1621}-5}{14} x=\frac{-\sqrt{1621}-5}{14}
The equation is now solved.
7x^{2}+5x-3=54
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}+5x-3-\left(-3\right)=54-\left(-3\right)
Add 3 to both sides of the equation.
7x^{2}+5x=54-\left(-3\right)
Subtracting -3 from itself leaves 0.
7x^{2}+5x=57
Subtract -3 from 54.
\frac{7x^{2}+5x}{7}=\frac{57}{7}
Divide both sides by 7.
x^{2}+\frac{5}{7}x=\frac{57}{7}
Dividing by 7 undoes the multiplication by 7.
x^{2}+\frac{5}{7}x+\left(\frac{5}{14}\right)^{2}=\frac{57}{7}+\left(\frac{5}{14}\right)^{2}
Divide \frac{5}{7}, the coefficient of the x term, by 2 to get \frac{5}{14}. Then add the square of \frac{5}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{7}x+\frac{25}{196}=\frac{57}{7}+\frac{25}{196}
Square \frac{5}{14} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{7}x+\frac{25}{196}=\frac{1621}{196}
Add \frac{57}{7} to \frac{25}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{14}\right)^{2}=\frac{1621}{196}
Factor x^{2}+\frac{5}{7}x+\frac{25}{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{5}{14}\right)^{2}}=\sqrt{\frac{1621}{196}}
Take the square root of both sides of the equation.
x+\frac{5}{14}=\frac{\sqrt{1621}}{14} x+\frac{5}{14}=-\frac{\sqrt{1621}}{14}
Simplify.
x=\frac{\sqrt{1621}-5}{14} x=\frac{-\sqrt{1621}-5}{14}
Subtract \frac{5}{14} from both sides of the equation.