You can solve quadric equations of the formax² + bx + c = 0Example: 2x² + 5x + 23 = 0with the following formula:x1 = (-b + sqrt(b² - 4*a*c)) / (2*a)x2 = (-b - sqrt(b² - 4*a*c)) / (2*a)where x1 and x2 are the two possible solutions of the equation.More details: http://en.wikipedia.org/wiki/Quadratic_equations
What is a quadratic equation?A quadratic equation is an equation that can be written in this form. ax2+bx+c=0 The a,b, and c here represent real number coefficients. So this means we are talking about an equation that is a constant times the variable squared plus a constant times the variable plus a constant equals zero, where the coefficient a on the variable squared can't be zero, because if it were then it would be a linear equation. Examples2x2+3x+1=0, x2+x=2x+3, (x+2)(x+3)=5 All these equations are equivalent to equations of the above form. The first one is already in that form. The second one can be put into it by subtracting 2x+3 from both sides. The third one can be put into it by multiplying out and then subtracting 5 from both sides.
The formula for a quadratic equation is as follows: AX^2+BX+C=0the formula to solve any quadratic equation is as follows: -b +or- the squareroot of b^2 - 4ac all over 2a. That is, if you put the equation in standard form such that it goes (a)+(b)+(c)=0. If and only if (a) is the term with the variable to the highest power, b is a number times a variable, and c is just a number.EX: A^2+3A-4 Normally you could not solve this equation by factoring. So you use the formula. (-3+Squareroot of(9-4*1*-4))/(2)This is an imaginary answer because the squareroot of -7 is (2.6457...*i)^2. the entire answer would be: ((-3+(2.6457...*i))/2) and ((-3-2.6457...*i))/2)
ax^2 + bx + c = 0-----------------delta= b^2 - 4acif delta > 0 then x1= (-b + Sqrt(delta))/2a x2= (-b - Sqrt(delta))/2a [x1 and x2 are two real numbers]if delta = 0 then x= -b/2a [x is a real number]if delta < 0 then x1= (-b + i*Sqrt(-delta))/2a x2= (-b - i*Sqrt(-delta))/2a i= sqrt(-1) [x1 and x2 are two complex numbers]