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xKingx
Can someone show me how I would solve this using the Quadratic Formula? 2x^2 – 16x + 32 = 0
do you know the quadratic form?
Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
The +- part always confuses me. Could you demonstrate how I'd substitute an equation like this into the form?
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ means:} \\ \quad x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \text{, and} \\ \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \text{ are the two solutions} \] Basically, its a more compact way of writing both solutions. The typical method is to substitute your values, simplify your sqrt(b^2 - 4ac) and -b, and then deal with splitting the \(\pm\) into two solutions + and -.
2x^2-8x-8x+32=0 2x(x-4)-8(x-4)=0 (2x-8)(x-4)=0 x=4,2
Can I ask why you wrote it as 2x^2-8x-8x+32=0
its supposed to be b+-4ac?