If it's not what You are looking for type in the equation solver your own equation and let us solve it.
4x^2-240x+576=0
a = 4; b = -240; c = +576;
Δ = b2-4ac
Δ = -2402-4·4·576
Δ = 48384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48384}=\sqrt{2304*21}=\sqrt{2304}*\sqrt{21}=48\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-48\sqrt{21}}{2*4}=\frac{240-48\sqrt{21}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+48\sqrt{21}}{2*4}=\frac{240+48\sqrt{21}}{8} $
| 2.27x+3.5=14.85 | | 10(y+6y)=10 | | 18x+3=11x-4 | | (3/5)x+(1/2)=(7/6) | | 4x+3=5(2x-8 | | 20x+14=54 | | 145=h=294 | | q-3.5=1.6 | | 11x^2-15x+4=0 | | t^2+3t-238=0 | | 2/3x=48/18+15/18-5/3 | | 18x-20=180 | | -12p+36=11(p+14) | | 5/3+2/3x=29/12+5/4x+1/4 | | 16x^2-4(x+1)(x+1)=0 | | -12p+36=11(p+14 | | h(11)=-20*11(11) | | (4x)^2-4(x+1)(x+1)=0 | | 125x=325x+1.2(30,000) | | -5-2n=4 | | 125x=325x+1.2x30,000 | | 4d=13 | | 9(n-3)=8+12 | | 6x-3(x-2)=2x+11 | | 4-2/3x=2x | | 234=y+189 | | x-246=122 | | 3/4-2/3+x=1/3 | | (6/x)+(7/6x)=44 | | X-2(2x-3)=4-x | | -7-30z+6=-15z-24-2 | | 4x^2-4(x+1)(x+1)=0 |