+4622 We can solve this using quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as \(x=\frac{-3+\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}, \quad \frac{-3+\sqrt{29}}{2}\) and the other solution as \(x=\frac{-3-\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}:\quad \frac{-3-\sqrt{29}}{2}.\) Thus, the final solutions as \(x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2}.\)
+4622 We can solve this using quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as \(x=\frac{-3+\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}, \quad \frac{-3+\sqrt{29}}{2}\) and the other solution as \(x=\frac{-3-\sqrt{3^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}:\quad \frac{-3-\sqrt{29}}{2}.\) Thus, the final solutions as \(x=\frac{-3+\sqrt{29}}{2},\:x=\frac{-3-\sqrt{29}}{2}.\)
+4622 Sorry if my explanation was confusing. Read more about the quadratic formula https://www.purplemath.com/modules/quadform.htm and https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/a/quadratic-formula-review, and I'm sure that you'll get a deeper understanding.
+129852 1x^2 + 3x - 5 = 0
We have the form ax^2 + bx + c = 0
Where a = 1 b = 3 and c = -5
The quadraic formula is
[ - b ± √ [ b^2 - (4 * a * c) ] ] / [ 2 * a ]
Plugging in for a, b, c......we have
[ - 3 ± √ [ 3^2 - (4 * 1 * -5) ] ] / [ 2 * 1 ] simplify
[ -3 ± √ [ 9 - ( -20) ] } / 2
[ - 3 ± √ 29 ] / 2
And the two solutions as are tertre found
x = [ -3 + √ 29 ] / 2 and x = [- 3 - √ 29 ] / 2
