I'd substitute the values -2, 8 and 130 into the quadratic formula and wind up with two real solutions: 2 plus or minus -1 times the square root of 69
x = -6.306623862918075
x = 10.306623862918075
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For a start, take out the common factor, 2. After that, you can try to factor the trinomial. If you have trouble with that, use the quadratic equation.
Pros: There are many real life situations in which the relationship between two variables is quadratic rather than linear. So to solve these situations quadratic equations are necessary. There is a simple equation to solve any quadratic equation. Cons: Pupils who are still studying basic mathematics will not be told how to solve quadratic equations in some circumstances - when the solutions lie in the Complex field.
You convert the equation to the form: ax2 + bx + c = 0, replace the numeric values (a, b, c) in the quadratic formula, and calculate.
If the discriminant of a quadratic equation is less than zero then it will not have any real roots.
y=±√15
For an equation of the form ax² + bx + c = 0 you can find the values of x that will satisfy the equation using the quadratic equation: x = [-b ± √(b² - 4ac)]/2a