A quadratic equation normally has 2 solutions and can be solved by using the quadratic equation formula.
2t2+8t+5 = 0 The expression in this quadratic equation is not so simple to factorise because 5 is a prime number which has only two factors (itself and one) but we can get a near enough solution by using the quadratic equation formula. Using the quadratic equation formula gives the solution as: t = - 0.7752551286 or t = - 3.224744871
By using the quadratic equation formula
By using the quadratic equation formula or by completing the square
To find the solution to this equation, you need to rearrange the terms and solve for the variable. 4 = 2b + b^2 can be rewritten as b^2 + 2b - 4 = 0. You can then solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Using the quadratic equation formula: x = -5-/+ the square root of 7
Without an equality sign the given expression can't be considered to be an equation but if it equals 0 then using the quadratic equation formula will give its solutions.
A quadratic equation normally has 2 solutions and can be solved by using the quadratic equation formula.
2t2+8t+5 = 0 The expression in this quadratic equation is not so simple to factorise because 5 is a prime number which has only two factors (itself and one) but we can get a near enough solution by using the quadratic equation formula. Using the quadratic equation formula gives the solution as: t = - 0.7752551286 or t = - 3.224744871
By using the quadratic equation formula
By using the quadratic equation formula or by completing the square
To find the solution to this equation, you need to rearrange the terms and solve for the variable. 4 = 2b + b^2 can be rewritten as b^2 + 2b - 4 = 0. You can then solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Yes. You can calculate the two roots of a quadratic equation by using the quadratic formula, and because there are square roots on the quadratic formula, and if the radicand is not a perfect square, so the answer to that equation has decimal.
It is a quadratic equation and its solutions can be found by using the quadratic equation formula.
The equation does not have a real number solution. Using the quadratic formula will give it's conjugate pair complex solution.
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
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