If it's not what You are looking for type in the equation solver your own equation and let us solve it.
4x^2-64x+232=0
a = 4; b = -64; c = +232;
Δ = b2-4ac
Δ = -642-4·4·232
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-8\sqrt{6}}{2*4}=\frac{64-8\sqrt{6}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+8\sqrt{6}}{2*4}=\frac{64+8\sqrt{6}}{8} $
| 8x4−42x3+29x2−42x+8=0 | | 8.x+x+x+x=64 | | 5.2x=80 | | (2p^2+4p)/(p^2+4p+3)=0 | | 1/2(16y-32)=3y-12+5y-4 | | 4^2=(x+3)3 | | 2x^2-27x-45=0 | | 5/8g-8=1/8g | | 5m−15=−5m= | | 12(16y-32)=3y-12+5y-4 | | 8y=54+2y | | 4y/5y=20 | | 20p-10=12p+54 | | ½h=15 | | 5z-2=8z+1 | | 12(x+3)=10(x) | | 8x^2=3–10x | | 2x^2+44x=0 | | 56+-3x=41 | | x^2-20x+82=-17 | | 9u=10.8 | | 9(p-1)=10(p-4) | | x^2-5x-78=6 | | x^2-5x-78=-6 | | 41-4x=13 | | x^2-5x-78=-15 | | c=1/4c-10 | | 113+-6x=53 | | x^2-17x+57=-15 | | 2^2-17x+57=-15 | | 8−2v=–5v−10 | | 8d+5=5+8d |