+4625 We can solve this using quadratic formula, x=−b±√b2−4ac2a . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as x=−3+√32−4⋅1(−5)2⋅1,−3+√292 and the other solution as x=−3−√32−4⋅1(−5)2⋅1:−3−√292. Thus, the final solutions as x=−3+√292,x=−3−√292.
+4625 We can solve this using quadratic formula, x=−b±√b2−4ac2a . (Glad that web2.0calc gives you this). Plugging the numbers into the quadratic formula, we have one solution as x=−3+√32−4⋅1(−5)2⋅1,−3+√292 and the other solution as x=−3−√32−4⋅1(−5)2⋅1:−3−√292. Thus, the final solutions as x=−3+√292,x=−3−√292.
+4625 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.
+130474 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
