If it's not what You are looking for type in the equation solver your own equation and let us solve it.
3=2x^2-4x
We move all terms to the left:
3-(2x^2-4x)=0
We get rid of parentheses
-2x^2+4x+3=0
a = -2; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·(-2)·3
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*-2}=\frac{-4-2\sqrt{10}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*-2}=\frac{-4+2\sqrt{10}}{-4} $
| 7=4v+43=2v | | -17=-5a-7 | | 1/2x+4=5/2x-8 | | 2x2+11x+21=-(x-5)(x+5) | | x^2+16-320=0 | | 7x+2+x+8=0 | | -55+6d+3=34 | | 2.1+10m=8.94 | | 0.325=39/b | | 5x-+2x-8=20 | | 10x-35=15x+40=4x+7 | | 31=-2.1h+1 | | (6+8)=(3x+38) | | 8x(7-3x)+1=4(2-x)(2+x) | | x3+x2-20=340 | | 14+y=3 | | 0.4y−8=9−0.6y | | 8(7-3x)+1=4(2-x)(2+x) | | |k|+2=10 | | |4x-1|=2x+9 | | x2+√2x-4=0 | | (7^x-2√10)(7^+2√10)=9 | | -0.6(3x-4)+3x=0.833 | | (5x+2)/3=8 | | 6=|s|−6 | | 23n+8=12 | | -1,5x2+3,5x=-3 | | 61=-2-7x | | -1/2x-8=2 | | |r|+6=10 | | 1,5x2+3,5x=-3 | | (7-4i)-(10-3i)=0 |