x = (- G - √D) /2 H and x = (- G + √D) /2 H or
Where, 'D' represents the determinant of quadratic equation and is equals to G2 – 4 H * K. So, our equation becomes:
x = (- G - √ (G2 – 4 H K)) /2 H and x = (- G + √(G2 – 4 H K)) /2 H............. equation (1).
If you want to know how do you know if a quadratic equation will have one, two, or no solutions before - hand, then we need to observe the value of determinant i.e. D.
In equation 1 if we substitute the value of D = 0 i.e.
G2 – 4 H * K = 0,
Or G = ± 2 √(H * K).
Then we are left with x = - G /2 H. So, in this situation we have only one root for the quadratic equation.
If 'D' exists and is greater than 0 i.e. G2 – 4 H * K > 0, then we get two roots for the equation: one positive and other one as negative i.e.
x = (- G ±√ (G2 – 4 H K)) /2 H,
If the value of D < 0 i.e. G2 – 4 H K < 0 then we get imaginary roots for quadratic equation i.e.
x = (- G ± i(G2 – 4 H K)) /2 H,
Where, i= √-1 and in such a situation there is no solution for quadratic equation.