If it's not what You are looking for type in the equation solver your own equation and let us solve it.
2x^2-42x+160=0
a = 2; b = -42; c = +160;
Δ = b2-4ac
Δ = -422-4·2·160
Δ = 484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-22}{2*2}=\frac{20}{4} =5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+22}{2*2}=\frac{64}{4} =16 $
| x²-11x-4=2x-18 | | B=-(2x+3)-2x | | x^2-42x+160=0 | | 5x+2=-10-3x-4 | | 2a+2=1 | | x^2+42x+160=0 | | -3-4=7y-14 | | 12(p−17)=192 | | 6v+v=3v+33 | | 6x-13=8x-5 | | x^2-(20*x)+64=0 | | 9x-6x+10=6 | | 17x÷5=3 | | 167–(7x+3)=108 | | 1.8*x=0.2 | | -3x+6=-6x-1 | | s·6+30=162 | | 5+6x=8x+11 | | y-3+4=23-2y | | 2699000=x+0.1x | | x^2-(42*x)+160=0 | | 1.25t^2+10t-30=0 | | x²-42*x+160=0 | | x²-42x+160=0 | | 23=4x^2-13 | | 12x(x)=132 | | 6y-3y=3y+5 | | (y)/(2)5(y+3)=(3)/(2)(4-3y) | | 3(4x-10)=2(15x-6) | | 10x+15=33+12x-58 | | 4t-6=24 | | X=2(5x-7) |